Chapter 3 · Motion in a Plane

3.3 Motion in a Plane with Constant Acceleration and Projectile Motion

Position, velocity and acceleration become vectors. With constant acceleration the x- and y-motions run independently, and a thrown ball follows a parabola whose range is $R=\dfrac{u^2\sin2\theta}{g}$.

Stand on a balcony and, at the same instant, drop one ball and throw another straight out horizontally. Which lands first? Most people guess the dropped ball — the thrown one "has further to go". In fact (ignoring air resistance) they hit the ground together. The horizontal throw does nothing at all to the falling: at every instant both balls are at the same height. Galileo realised this four centuries ago, and it is the key to every thrown, kicked, hit or launched object — a cricket ball, a long jumper, a water jet, a parcel dropped from a plane.

In Chapter 2 you described motion along a line with $x$, $v$ and $a$. In a plane the same quantities become vectors: the position $\mathbf{r}$, the velocity $\mathbf{v}$ and the acceleration $\mathbf{a}$. When the acceleration is constant, the motion splits into two straight-line motions — one along $x$, one along $y$ — that share only the clock. Projectile motion is the most important example: constant horizontal velocity plus free fall. From that one idea we will derive the shape of the path (a parabola), the time of flight, the greatest height, and the range — and see why $45^\circ$ throws go furthest.

Position, velocity and acceleration in a plane

In a plane, the position of a point object is its position vector from the origin, $\mathbf{r}=x\,\hat{\mathbf{i}}+y\,\hat{\mathbf{j}}$. As it moves, $x$ and $y$ both change with time.

📘 Definition 3.7 — Displacement and velocity in a plane

If the object moves from $\mathbf{r}$ to $\mathbf{r}'$ in a time $\Delta t$, its displacement is $\Delta\mathbf{r}=\mathbf{r}'-\mathbf{r}=\Delta x\,\hat{\mathbf{i}}+\Delta y\,\hat{\mathbf{j}}$ and its average velocity is $\bar{\mathbf{v}}=\dfrac{\Delta\mathbf{r}}{\Delta t}$, a vector along $\Delta\mathbf{r}$.

The (instantaneous) velocity is the limit as $\Delta t\to0$:

$$\mathbf{v}=\lim_{\Delta t\to0}\frac{\Delta\mathbf{r}}{\Delta t}=\frac{d\mathbf{r}}{dt}=v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}},\qquad v_x=\frac{dx}{dt},\quad v_y=\frac{dy}{dt}.$$

The speed is $v=\sqrt{v_x^2+v_y^2}$ and the direction of motion makes an angle $\theta$ with $+x$ where $\tan\theta=\dfrac{v_y}{v_x}$. As $\Delta t\to0$ the chord $\Delta\mathbf{r}$ turns into the tangent, so the velocity is always tangent to the path, pointing the way the object is going.

📘 Definition 3.8 — Acceleration in a plane

The average acceleration over $\Delta t$ is $\bar{\mathbf{a}}=\dfrac{\Delta\mathbf{v}}{\Delta t}$ and the (instantaneous) acceleration is

$$\mathbf{a}=\frac{d\mathbf{v}}{dt}=a_x\,\hat{\mathbf{i}}+a_y\,\hat{\mathbf{j}},\qquad a_x=\frac{dv_x}{dt},\quad a_y=\frac{dv_y}{dt}.$$

In a plane, $\mathbf{a}$ need not point along $\mathbf{v}$. A change in the direction of the velocity is an acceleration just as much as a change in speed (Example 6 of Section 3.1).

So each component behaves exactly like a Chapter 2 velocity or acceleration: $v_x$ is the slope of the $x$–$t$ graph, $a_y$ is the slope of the $v_y$–$t$ graph, and so on. Only at the end do we put the components back together into a vector.

✏️ Example 1 — A drone's curved flight

A drone's position is $x=6.0\,t$, $\ y=8.0\,t-1.0\,t^2$ (in metres, $t$ in seconds). Find (a) its velocity and acceleration at any time, (b) its velocity at $t=2.0\ \text{s}$ (magnitude and direction), (c) when and where its velocity is horizontal, (d) the equation of its path.

💡 Show solution▶

(a) $v_x=\dfrac{dx}{dt}=6.0\ \text{m/s}$, $\ v_y=\dfrac{dy}{dt}=8.0-2.0\,t$; $\ a_x=0$, $\ a_y=\dfrac{dv_y}{dt}=-2.0\ \text{m/s}^2$. So $\mathbf{v}=6.0\,\hat{\mathbf{i}}+(8.0-2.0t)\,\hat{\mathbf{j}}$ and $\mathbf{a}=-2.0\,\hat{\mathbf{j}}\ \text{m/s}^2$, constant.

(b) $\mathbf{v}=6.0\,\hat{\mathbf{i}}+4.0\,\hat{\mathbf{j}}$: $v=\sqrt{36+16}=7.2\ \text{m/s}$ at $\tan^{-1}\tfrac{4.0}{6.0}=33.7^\circ$ above $+x$.

(d) $t=\dfrac{x}{6.0}$, so $y=\dfrac{8.0}{6.0}x-\dfrac{x^2}{36}$: a parabola, because $\mathbf{a}$ is constant and not along $\mathbf{v}$.

Check At $x=24$: $y=32-16=16\ \text{m}$ ✓.

Motion in a plane with constant acceleration

If $\mathbf{a}$ is constant, then $a_x$ and $a_y$ are both constant, and each coordinate obeys the kinematic equations of Chapter 2 on its own. Writing them side by side and recombining gives vector equations.

📐 Result 3.7 — Constant acceleration in a plane

If a body has constant acceleration $\mathbf{a}$, and position $\mathbf{r}_0$ and velocity $\mathbf{v}_0$ at $t=0$, then at time $t$

$$\mathbf{v}=\mathbf{v}_0+\mathbf{a}\,t,\qquad \mathbf{r}=\mathbf{r}_0+\mathbf{v}_0\,t+\tfrac12\mathbf{a}\,t^2 .$$

In components:

$$v_x=v_{0x}+a_xt,\qquad x=x_0+v_{0x}t+\tfrac12a_xt^2,$$ $$v_y=v_{0y}+a_yt,\qquad y=y_0+v_{0y}t+\tfrac12a_yt^2.$$

The motion is two independent straight-line motions with constant acceleration, one along $x$ and one along $y$, linked only by the common time $t$. (These hold only if $\mathbf{a}$ is constant in magnitude and direction.)

The calculus derivation is the one from Chapter 2, done once for each component: the velocity whose rate of change is the constant $a_x$ is $v_{0x}+a_xt$, and the position whose rate of change is that velocity is $x_0+v_{0x}t+\tfrac12a_xt^2$. Graphically, the area under the $v_x$–$t$ line (a trapezium) is the change in $x$.

✏️ Example 2 — A hovercraft in a steady wind

A model hovercraft leaves a buoy (the origin) with velocity $\mathbf{v}_0=(3.0\,\hat{\mathbf{i}}+4.0\,\hat{\mathbf{j}})\ \text{m/s}$. A steady wind gives it a constant acceleration $\mathbf{a}=-0.50\,\hat{\mathbf{i}}\ \text{m/s}^2$. Find (a) its velocity and position at $t=4.0\ \text{s}$, (b) when and where it crosses the $y$-axis again.

💡 Show solution▶

x-motion $v_x=3.0-0.50t$, $\ x=3.0t-0.25t^2$. y-motion (no acceleration): $v_y=4.0$, $\ y=4.0t$.

(a) At $t=4.0\ \text{s}$: $\mathbf{v}=(1.0\,\hat{\mathbf{i}}+4.0\,\hat{\mathbf{j}})\ \text{m/s}$ (speed $4.1\ \text{m/s}$); $\mathbf{r}=(12-4.0)\,\hat{\mathbf{i}}+16\,\hat{\mathbf{j}}=(8.0\,\hat{\mathbf{i}}+16\,\hat{\mathbf{j}})\ \text{m}$.

(b) $x=0$: $t(3.0-0.25t)=0$, so $t=12\ \text{s}$; then $y=4.0\times12=48\ \text{m}$. It crosses at $(0,\ 48\ \text{m})$.

Check The $x$-motion alone is a Chapter 2 "throw up and come back" with $u=3.0$, $a=-0.50$: back to $x=0$ after $2u/|a|=12\ \text{s}$ ✓.

At $t=0$ one ball is dropped and another is thrown horizontally from the same height. Faint images are taken every $0.20\ \text{s}$ (a strobe photograph). Watch the dashed lines: both balls are at the same height at every flash, and the thrown ball moves equal distances sideways in equal times. On the thrown ball, red is $v_x$ (constant), blue is $v_y$ (growing) and purple is $\mathbf{v}$, always along the path. Air resistance is neglected; $g=9.8\ \text{m/s}^2$.

height $h$:20 mthrow speed $u$:15 m/s

time to land (both)—

horizontal distance—

$v_y$ on landing—

landing velocity—

Projectile motion

📘 Definition 3.9 — Projectile

A projectile is a body that, after being launched, moves under gravity alone. Near the Earth's surface, with air resistance neglected, its acceleration is constant: $\mathbf{a}=-g\,\hat{\mathbf{j}}$ ($x$ horizontal, $y$ vertically up, $g=9.8\ \text{m/s}^2$). Its path is called its trajectory. If it is launched from the origin with speed $u$ at an angle of projection $\theta$ above the horizontal, then $u_x=u\cos\theta$ and $u_y=u\sin\theta$.

Put $a_x=0$ and $a_y=-g$ into Result 3.7. Horizontally nothing pushes the body, so its horizontal velocity never changes; vertically it is in free fall (Chapter 2) with initial velocity $u\sin\theta$. Eliminating $t=\dfrac{x}{u\cos\theta}$ from the $y$-equation gives the shape of the path.

📐 Result 3.8 — Equations of a projectile

For a projectile launched from the origin with speed $u$ at angle $\theta$ above the horizontal (constant $g$, no air resistance):

$$v_x=u\cos\theta\ \text{(constant)},\qquad x=(u\cos\theta)\,t,$$ $$v_y=u\sin\theta-gt,\qquad y=(u\sin\theta)\,t-\tfrac12gt^2 .$$

The trajectory is the parabola

$$y=x\tan\theta-\frac{g\,x^2}{2u^2\cos^2\theta}.$$

Three numbers describe a flight that starts and ends on level ground. Time to the top: at the highest point $v_y=0$, so $t_m=\dfrac{u\sin\theta}{g}$. Time of flight: $y=0$ again when $t\,(u\sin\theta-\tfrac12gt)=0$, at $T=\dfrac{2u\sin\theta}{g}=2t_m$ — the way up and the way down take equal times. Maximum height: put $t_m$ into $y$, or use $v_y^2=u_y^2-2gy$ with $v_y=0$. Range: $R=(u\cos\theta)\,T=\dfrac{2u^2\sin\theta\cos\theta}{g}$, and $2\sin\theta\cos\theta=\sin2\theta$ (you will prove this identity in maths).

📐 Result 3.9 — Time of flight, maximum height and range

For a projectile launched and landing at the same level (constant $g$, no air resistance):

$$T=\frac{2u\sin\theta}{g},\qquad H=\frac{u^2\sin^2\theta}{2g},\qquad R=\frac{u^2\sin2\theta}{g}.$$

The range is greatest when $\sin2\theta=1$, i.e. at $\theta=45^\circ$, where $R_\text{max}=\dfrac{u^2}{g}$. Since $\sin2\theta=\sin(180^\circ-2\theta)$, the complementary angles $\theta$ and $90^\circ-\theta$ give the same range (the steeper one flies higher and longer).

✏️ Example 3 — Kicked off a cliff

A stone is kicked horizontally at $15\ \text{m/s}$ from the top of a cliff $20\ \text{m}$ high. Find (a) the time it takes to reach the ground below, (b) how far from the foot of the cliff it lands, (c) its velocity just before it lands.

💡 Show solution▶

Set up $\theta=0$: $u_x=15\ \text{m/s}$, $u_y=0$. Take $y$ downward for the fall: $h=\tfrac12gt^2$.

(a) $t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\times20}{9.8}}=2.02\ \text{s}$ — exactly the time to drop from $20\ \text{m}$.

(b) $x=u_xt=15\times2.02=30\ \text{m}$ (more precisely $30.3\ \text{m}$).

(c) $v_x=15\ \text{m/s}$, $v_y=gt=19.8\ \text{m/s}$ downward. $v=\sqrt{15^2+19.8^2}=24.8\ \text{m/s}$ at $\tan^{-1}\tfrac{19.8}{15}=52.9^\circ$ below the horizontal.

Check Energy (Chapter 5) agrees: $v^2=u^2+2gh=225+392=617$, $v=24.8\ \text{m/s}$ ✓. These are the default settings of Interactive 1.

✏️ Example 4 — A lofted cricket shot

A batter lofts a cricket ball at $25\ \text{m/s}$ at $40^\circ$ above the horizontal. Treating the ball as launched and caught at ground level, find (a) the time of flight, (b) the greatest height, (c) the horizontal range, (d) the ball's velocity at the top.

💡 Show solution▶

Components $u_x=25\cos40^\circ=19.2\ \text{m/s}$, $u_y=25\sin40^\circ=16.1\ \text{m/s}$.

(a) $T=\dfrac{2u_y}{g}=\dfrac{2\times16.07}{9.8}=3.28\ \text{s}$.

(b), (c) $H=\dfrac{u_y^2}{2g}=\dfrac{16.07^2}{19.6}=13.2\ \text{m}$; $\ R=\dfrac{u^2\sin80^\circ}{g}=\dfrac{625\times0.985}{9.8}=62.8\ \text{m}$.

(d) At the top $v_y=0$: the velocity is $19.2\ \text{m/s}$ horizontally — not zero.

Check $R=u_xT=19.15\times3.28=62.8\ \text{m}$ ✓. Set $u=25$, $\theta=40^\circ$ in Interactive 2 to see the same numbers.

✏️ Example 5 — Two angles, one range

A hose sprays water at $30\ \text{m/s}$ from ground level. (a) What is the greatest distance it can reach on level ground? (b) At what two angles will it reach a point $60\ \text{m}$ away? (c) Which of the two jets arrives first?

💡 Show solution▶

(a) $R_\text{max}=\dfrac{u^2}{g}=\dfrac{900}{9.8}=91.8\ \text{m}$, at $45^\circ$.

(b) $\sin2\theta=\dfrac{Rg}{u^2}=\dfrac{60\times9.8}{900}=0.653$, so $2\theta=40.8^\circ$ or $180^\circ-40.8^\circ=139.2^\circ$: $\theta=20.4^\circ$ or $69.6^\circ$ — complementary, adding to $90^\circ$.

(c) $T=\dfrac{2u\sin\theta}{g}$: $2.13\ \text{s}$ at $20.4^\circ$ and $5.74\ \text{s}$ at $69.6^\circ$. The low jet arrives first.

Check $T_1T_2=\dfrac{4u^2\sin\theta\cos\theta}{g^2}=\dfrac{2R}{g}=12.2\ \text{s}^2$, and $2.13\times5.74=12.2$ ✓.

✏️ Example 6 — Under the crossbar?

A footballer strikes a free kick at $20\ \text{m/s}$ at $25^\circ$ above the horizontal, $25\ \text{m}$ from the goal line. The crossbar is $2.44\ \text{m}$ high. Neglecting spin and air resistance, does the ball pass under the bar? Is it rising or falling as it crosses the line?

💡 Show solution▶

Trajectory $y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta}$ with $x=25\ \text{m}$: $\ y=25(0.4663)-\dfrac{9.8\times625}{2\times400\times0.8214}=11.66-9.32=2.34\ \text{m}$.

Verdict $2.34\ \text{m}\lt2.44\ \text{m}$: it passes about $10\ \text{cm}$ under the bar.

Rising? It reaches the line at $t=\dfrac{25}{20\cos25^\circ}=1.38\ \text{s}$, when $v_y=20\sin25^\circ-9.8\times1.38=-5.1\ \text{m/s}$: falling, dipping under the bar.

Check The top of the flight is at $t_m=\tfrac{8.45}{9.8}=0.86\ \text{s}$, before $1.38\ \text{s}$ — consistent with "falling" ✓.

Choose the launch speed, the angle and the planet, then press Play. The ball traces its parabola computed from $x=(u\cos\theta)t$, $y=(u\sin\theta)t-\tfrac12gt^2$; on the ball, red is $v_x$ (it never changes), blue is $v_y$ (shrinking, zero at the top, then growing downward) and purple is $\mathbf{v}$. The dashed purple path is the $45^\circ$ throw (greatest range); the dashed green path is the complementary angle $90^\circ-\theta$, which lands at the same spot.

speed $u$:25 m/sangle $\theta$:40°planet:

time of flight $T$—

max height $H$—

range $R$—

$R_\text{max}$ (at $45^\circ$)—

⚠️ Common mistakes

1. "At the top the velocity is zero." Only $v_y$ is zero; $v_x=u\cos\theta$ is still there. And the acceleration at the top is still $g$ downward — it is never zero in flight.

2. Using $R=u^2\sin2\theta/g$ when the levels differ. Result 3.9 assumes landing at the launch height. From a cliff or a balcony, use the component equations of Result 3.8 (Example 3 and Exercise IV.2).

3. Mixing up horizontal and vertical. Gravity enters only the $y$-equations; never put $g$ into $x=(u\cos\theta)t$.

4. Angle from the vertical. $\theta$ in these formulas is measured from the horizontal. If an angle is given from the vertical, convert it first.

5. Forgetting the second angle. $\sin2\theta=k$ has two solutions between $0^\circ$ and $90^\circ$ (unless $k=1$); both are valid launch angles.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — relative velocity in two dimensions

Relative velocity in a plane is no longer in the NCERT/CBSE syllabus, but JEE Main, JEE Advanced and NEET still ask about it. It needs nothing beyond vector subtraction.

Definition. If bodies $A$ and $B$ have velocities $\mathbf{v}_A$ and $\mathbf{v}_B$ relative to the ground, the velocity of $A$ relative to $B$ (as seen by an observer moving with $B$) is $\mathbf{v}_{AB}=\mathbf{v}_A-\mathbf{v}_B$, and $\mathbf{v}_{BA}=-\mathbf{v}_{AB}$.

Rain and umbrella. Rain falls vertically at $6.0\ \text{m/s}$; a cyclist rides east at $8.0\ \text{m/s}$. Relative to her, $\mathbf{v}_{\text{rain},\text{cyclist}}=\mathbf{v}_\text{rain}-\mathbf{v}_\text{cyclist}$ = $6.0\ \text{m/s}$ down plus $8.0\ \text{m/s}$ west: magnitude $\sqrt{36+64}=10\ \text{m/s}$, at $\tan^{-1}\tfrac{8.0}{6.0}=53^\circ$ to the vertical, coming at her from the front. She should tilt her umbrella $53^\circ$ forward.

Crossing a river. A river of width $d$ flows at $v_r$; a boat moves at $v_b$ relative to the water. (i) Shortest time: head straight across; $t=\dfrac{d}{v_b}$ and the current carries the boat a distance $v_r t$ downstream. (ii) No drift (possible only if $v_b\gt v_r$): head upstream at angle $\alpha$ to the straight-across line with $\sin\alpha=\dfrac{v_r}{v_b}$; then the speed across is $\sqrt{v_b^2-v_r^2}$ and $t=\dfrac{d}{\sqrt{v_b^2-v_r^2}}$. For $d=120\ \text{m}$, $v_r=3.0\ \text{m/s}$, $v_b=5.0\ \text{m/s}$: (i) $t=24\ \text{s}$ with $72\ \text{m}$ drift; (ii) $\alpha=36.9^\circ$ upstream and $t=\tfrac{120}{4.0}=30\ \text{s}$.

Exercises

Take $g=9.8\ \text{m/s}^2$ and neglect air resistance unless told otherwise.

Velocity and acceleration in a plane. A particle has position $\mathbf{r}=(3.0\,t)\,\hat{\mathbf{i}}+(2.0\,t^2-1.0)\,\hat{\mathbf{j}}$ (metres, seconds).

1 Find its velocity at any time $t$.
2 Find its acceleration. Is it constant?
3 Find its speed and direction of motion at $t=1.0\ \text{s}$.
4 Find the equation of its path, $y$ as a function of $x$.
5 A different particle's velocity changes from $4.0\,\hat{\mathbf{i}}\ \text{m/s}$ to $3.0\,\hat{\mathbf{j}}\ \text{m/s}$ in $2.0\ \text{s}$. Find its average acceleration (vector and magnitude).
6 A body starts at the origin with $\mathbf{v}_0=2.0\,\hat{\mathbf{i}}\ \text{m/s}$ and constant acceleration $\mathbf{a}=(1.0\,\hat{\mathbf{i}}+0.50\,\hat{\mathbf{j}})\ \text{m/s}^2$. Find its position and velocity at $t=4.0\ \text{s}$.
7 In part 6, when is $y=9.0\ \text{m}$, and what is $x$ then?

💡 Show solutions▶

1 $\mathbf{v}=\dfrac{d\mathbf{r}}{dt}=3.0\,\hat{\mathbf{i}}+4.0\,t\,\hat{\mathbf{j}}$ m/s.

2 $\mathbf{a}=4.0\,\hat{\mathbf{j}}\ \text{m/s}^2$: constant (magnitude and direction).

3 $\mathbf{v}=3.0\,\hat{\mathbf{i}}+4.0\,\hat{\mathbf{j}}$: $5.0\ \text{m/s}$ at $\tan^{-1}\tfrac43=53.1^\circ$ above $+x$.

4 $t=x/3.0$: $y=\tfrac{2.0}{9.0}x^2-1.0$, a parabola.

5 $\bar{\mathbf{a}}=\dfrac{3.0\,\hat{\mathbf{j}}-4.0\,\hat{\mathbf{i}}}{2.0}=(-2.0\,\hat{\mathbf{i}}+1.5\,\hat{\mathbf{j}})\ \text{m/s}^2$; magnitude $2.5\ \text{m/s}^2$.

6 $x=2.0(4.0)+\tfrac12(1.0)(16)=16\ \text{m}$, $y=\tfrac12(0.50)(16)=4.0\ \text{m}$; $\mathbf{v}=(2.0+4.0)\,\hat{\mathbf{i}}+(0.50\times4.0)\,\hat{\mathbf{j}}=(6.0\,\hat{\mathbf{i}}+2.0\,\hat{\mathbf{j}})\ \text{m/s}$.

7 $0.25t^2=9.0$, $t=6.0\ \text{s}$; $x=12+18=30\ \text{m}$.

Horizontal projection.

1 A marble rolls off a table $1.25\ \text{m}$ high at $2.0\ \text{m/s}$. How long is it in the air, and how far from the foot of the table does it land?
2 Find the marble's speed and direction just before it lands.
3 A relief plane flying horizontally at $80\ \text{m/s}$, $490\ \text{m}$ above flat ground, releases a supply packet. How long does the packet fall, how far ahead of the release point does it land, and where is the plane at that moment?
4 From the same height, one ball is dropped and another is thrown horizontally at $10\ \text{m/s}$ at the same instant. Which reaches the ground first? Explain.
5 A dart is thrown horizontally at $12\ \text{m/s}$, aimed straight at the centre of a board $3.0\ \text{m}$ away. How far below the centre does it hit?
6 A stone thrown horizontally from the top of a $45\ \text{m}$ tower lands $60\ \text{m}$ from its foot. Find the speed of throwing.
7 Show that a body thrown horizontally with speed $u$ from the origin follows $y=-\dfrac{g x^2}{2u^2}$ (with $y$ up). What shape is this?

💡 Show solutions▶

1 $t=\sqrt{2(1.25)/9.8}=0.505\ \text{s}$; $x=2.0\times0.505=1.0\ \text{m}$.

2 $v_y=9.8\times0.505=4.95\ \text{m/s}$; $v=\sqrt{2.0^2+4.95^2}=5.3\ \text{m/s}$ at $\tan^{-1}\tfrac{4.95}{2.0}=68^\circ$ below the horizontal.

3 $t=\sqrt{2(490)/9.8}=10\ \text{s}$; $x=80\times10=800\ \text{m}$. The plane is directly above the packet: both keep the same horizontal velocity.

4 They land together: the vertical motion ($u_y=0$, acceleration $g$) is the same for both; the horizontal velocity does not affect it.

5 $t=3.0/12=0.25\ \text{s}$; drop $=\tfrac12(9.8)(0.25)^2=0.31\ \text{m}$ below the centre.

6 $t=\sqrt{2(45)/9.8}=3.03\ \text{s}$; $u=60/3.03=19.8\ \text{m/s}$.

7 $x=ut$ gives $t=x/u$; $y=-\tfrac12gt^2=-\dfrac{gx^2}{2u^2}$: a downward-opening parabola with its vertex at the launch point (the $\theta=0$ case of Result 3.8).

III

Projectiles on level ground.

1 A ball is thrown at $20\ \text{m/s}$ at $30^\circ$ above the horizontal. Find its time of flight, maximum height and range.
2 Repeat part 1 for $60^\circ$. Which quantities change, and which do not?
3 What is the greatest range possible with a launch speed of $20\ \text{m/s}$?
4 What launch speed gives a maximum range of $50\ \text{m}$?
5 For what angle of projection is the maximum height equal to the range?
6 At the top of its flight a projectile's speed is half its launch speed. Find the angle of projection.
7 A ball is launched at $30\ \text{m/s}$ at an angle $\theta$ with $\tan\theta=\tfrac43$ (so $u_x=18\ \text{m/s}$, $u_y=24\ \text{m/s}$). Find its velocity (magnitude and direction) at $t=1.0\ \text{s}$ and at $t=3.0\ \text{s}$.
8 A javelin is released at $27\ \text{m/s}$ at $38^\circ$. Ignoring the release height, estimate the length of the throw.
9 The same throw as in part 8 is made on the Moon ($g=1.62\ \text{m/s}^2$). How far does it go?

💡 Show solutions▶

1 $T=\dfrac{2(20)(0.5)}{9.8}=2.04\ \text{s}$; $H=\dfrac{(10)^2}{19.6}=5.10\ \text{m}$; $R=\dfrac{400\sin60^\circ}{9.8}=35.3\ \text{m}$.

2 $T=3.53\ \text{s}$, $H=15.3\ \text{m}$, $R=35.3\ \text{m}$: the range is unchanged ($30^\circ$ and $60^\circ$ are complementary); the steeper throw is three times as high and $\sqrt3$ times as long in the air.

3 $R_\text{max}=\dfrac{400}{9.8}=40.8\ \text{m}$ at $45^\circ$.

4 $u=\sqrt{gR_\text{max}}=\sqrt{9.8\times50}=22.1\ \text{m/s}$.

5 $\dfrac{u^2\sin^2\theta}{2g}=\dfrac{2u^2\sin\theta\cos\theta}{g}$ gives $\tan\theta=4$: $\theta=76.0^\circ$.

6 At the top $v=u\cos\theta=\tfrac12u$: $\cos\theta=\tfrac12$, $\theta=60^\circ$.

7 $t=1.0\ \text{s}$: $v_y=24-9.8=14.2\ \text{m/s}$; $v=\sqrt{18^2+14.2^2}=22.9\ \text{m/s}$ at $38.3^\circ$ above the horizontal. $t=3.0\ \text{s}$: $v_y=24-29.4=-5.4\ \text{m/s}$; $v=18.8\ \text{m/s}$ at $16.7^\circ$ below the horizontal (it has passed the top at $2.45\ \text{s}$).

8 $R=\dfrac{27^2\sin76^\circ}{9.8}=72\ \text{m}$ (world-class throws are near $90\ \text{m}$: faster release and some aerodynamic lift).

9 $R\propto\dfrac1g$: $72.2\times\dfrac{9.8}{1.62}=437\ \text{m}$ — about six times as far.

Applications and harder problems.

1 A basketball player releases the ball $2.0\ \text{m}$ above the floor at $7.6\ \text{m/s}$ at $52^\circ$ above the horizontal. The centre of the hoop is $4.6\ \text{m}$ away horizontally and $3.05\ \text{m}$ above the floor. Is the ball above the rim height when it is over the centre of the hoop, and is it rising or falling?
2 A ball is thrown from the top of a $30\ \text{m}$ cliff at $15\ \text{m/s}$, $30^\circ$ above the horizontal. When and how far from the foot of the cliff does it land?
3 A long jumper takes off at $9.5\ \text{m/s}$ at $20^\circ$ above the horizontal. Treating her as a point that lands at take-off height, how long is the jump?
4 Water leaves a garden hose at $12\ \text{m/s}$. How far can it reach on level ground, and at what two angles does it reach a flower bed $10\ \text{m}$ away?
5 Show that when a projectile is launched for maximum range, its maximum height is one quarter of that range.
6 (Modelling) A fielder can throw at $30\ \text{m/s}$. Can he reach the wicket-keeper $80\ \text{m}$ away without a bounce? If so, at what angles, and why do fielders prefer the lower one? Real throws travel less far than this model predicts — suggest why.

💡 Show solutions▶

1 $u_x=7.6\cos52^\circ=4.68\ \text{m/s}$, so $t=\tfrac{4.6}{4.68}=0.983\ \text{s}$. Height $=2.0+(7.6\sin52^\circ)(0.983)-4.9(0.983)^2=2.0+5.887-4.735=3.15\ \text{m}$, about $10\ \text{cm}$ above the rim; $v_y=5.99-9.63=-3.6\ \text{m/s}$, so it is falling — it drops into the basket.

2 With $y$ up from the cliff top: $-30=7.5t-4.9t^2$, i.e. $4.9t^2-7.5t-30=0$, $t=\dfrac{7.5+\sqrt{56.25+588}}{9.8}=3.36\ \text{s}$ (the negative root is rejected). $x=(15\cos30^\circ)(3.36)=43.6\ \text{m}$.

3 $R=\dfrac{9.5^2\sin40^\circ}{9.8}=5.9\ \text{m}$.

4 $R_\text{max}=\tfrac{144}{9.8}=14.7\ \text{m}$. $\sin2\theta=\tfrac{10\times9.8}{144}=0.681$: $2\theta=42.9^\circ$ or $137.1^\circ$, so $\theta=21.4^\circ$ or $68.6^\circ$.

5 At $45^\circ$: $R=\dfrac{u^2}{g}$ and $H=\dfrac{u^2\sin^245^\circ}{2g}=\dfrac{u^2}{4g}=\dfrac R4$.

6 $R_\text{max}=\tfrac{900}{9.8}=91.8\ \text{m}\gt80\ \text{m}$, so yes. $\sin2\theta=\tfrac{80\times9.8}{900}=0.871$: $\theta=30.3^\circ$ or $59.7^\circ$. Flight times $3.09\ \text{s}$ and $5.29\ \text{s}$: the low throw arrives over $2\ \text{s}$ sooner, which wins run-outs. Air resistance slows a real ball, so it falls short of the model's range.

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 5 marks A projectile is fired from level ground with speed $u$ at angle $\theta$ to the horizontal. Show that its path is a parabola, and derive expressions for (i) the time of flight, (ii) the maximum height, (iii) the horizontal range. Show that the range is greatest for $\theta=45^\circ$.

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Path $a_x=0$, $a_y=-g$: $x=(u\cos\theta)t$, $y=(u\sin\theta)t-\tfrac12gt^2$. Eliminating $t$: $y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta}$, of the form $y=bx-cx^2$ — a parabola. [1]

(i) $y=0$ at $t=\dfrac{2u\sin\theta}{g}$. [1]

(ii) At the top $v_y=0$: $0=u^2\sin^2\theta-2gH$, $H=\dfrac{u^2\sin^2\theta}{2g}$. [1]

(iii) $R=(u\cos\theta)T=\dfrac{u^2\sin2\theta}{g}$. [1]

Maximum $\sin2\theta\le1$ with equality at $2\theta=90^\circ$: $R_\text{max}=\dfrac{u^2}{g}$ at $\theta=45^\circ$. [1]

Exam tip State the assumptions (no air resistance, constant $g$, same launch and landing level) — they earn marks and show you know when the formulas fail.

E2. NEET At the highest point of the path of a projectile (no air resistance), (a) its velocity and acceleration are both zero (b) its velocity is horizontal and its acceleration is $g$ downward (c) its velocity is zero and its acceleration is $g$ downward (d) its velocity is horizontal and its acceleration is zero

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Idea $v_y=0$ but $v_x=u\cos\theta$ remains; gravity acts throughout.

Answer (b).

Exam tip Velocity and acceleration are perpendicular at the top — the only point where that happens in projectile motion.

E3. JEE Main Two balls are thrown with the same speed at $15^\circ$ and $75^\circ$ to the horizontal. The ratio of their maximum heights $H_{15}:H_{75}$ is (a) $1$ (b) $7-4\sqrt3$ (c) $2-\sqrt3$ (d) $\tfrac13$

💡 Show solution▶

Ratio $\dfrac{H_{15}}{H_{75}}=\dfrac{\sin^215^\circ}{\sin^275^\circ}=\tan^215^\circ$, since $\sin75^\circ=\cos15^\circ$.

Value $\tan15^\circ=2-\sqrt3$, so the ratio is $(2-\sqrt3)^2=7-4\sqrt3\approx0.072$.

Answer (b).

Exam tip (a) confuses height with range (equal for complementary angles); (c) forgets to square. Numerically, $\sin^2 15^\circ/\sin^2 75^\circ=0.0670/0.933=0.072$ — check the surds with a calculator.

E4. JEE Main (Numerical value) A ball is thrown from the top of a building $32\ \text{m}$ high with speed $20\ \text{m/s}$ at $37^\circ$ above the horizontal. Taking $g=10\ \text{m/s}^2$, $\sin37^\circ=0.6$ and $\cos37^\circ=0.8$, find the horizontal distance (in m) from the foot of the building to where the ball lands.

💡 Show solution▶

Components $u_x=16\ \text{m/s}$, $u_y=12\ \text{m/s}$.

Vertical $-32=12t-5t^2$, i.e. $5t^2-12t-32=0$, $t=\dfrac{12+\sqrt{144+640}}{10}=\dfrac{12+28}{10}=4.0\ \text{s}$.

Answer $x=16\times4.0=64\ \text{m}$.

Exam tip Never use $R=u^2\sin2\theta/g$ here ($=38.4\ \text{m}$, a tempting wrong answer): launch and landing levels differ.

E5. JEE Advanced (One or more correct) A projectile is launched from level ground with speed $u$ at angle $\theta$; another with the same speed at $90^\circ-\theta$. Neglect air resistance. Which statements are correct? (A) For each, the time to reach the top is half the time of flight. (B) Each lands with the same velocity it was launched with. (C) The sum of their maximum heights is $\dfrac{u^2}{2g}$. (D) The product of their times of flight is $\dfrac{2R}{g}$, where $R$ is their common range.

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(A) $t_m=\dfrac{u\sin\theta}{g}=\tfrac12T$ ✓ (symmetric flight on level ground).

(B) Same speed, but $v_y$ has reversed: the landing velocity points $\theta$ below the horizontal ✗.

(D) $T_1T_2=\dfrac{2u\sin\theta}{g}\cdot\dfrac{2u\cos\theta}{g}=\dfrac{2}{g}\cdot\dfrac{u^2\sin2\theta}{g}=\dfrac{2R}{g}$ ✓.

Answer (A), (C), (D). Example 5 checked (D) numerically.

E6. NEET JEE Main Rain is falling vertically at $12\ \text{m/s}$. A cyclist rides due east at $5.0\ \text{m/s}$. To keep dry she should hold her umbrella tilted (a) $22.6^\circ$ from the vertical towards the east (b) $22.6^\circ$ from the vertical towards the west (c) $67.4^\circ$ from the vertical towards the east (d) vertically

💡 Show solution▶

Note Relative velocity in two dimensions is beyond the NCERT syllabus (see the 🚀 box), but it is examined in JEE and NEET.

Relative velocity $\mathbf{v}_\text{rain,cyclist}=\mathbf{v}_\text{rain}-\mathbf{v}_\text{cyclist}$: $12\ \text{m/s}$ down and $5.0\ \text{m/s}$ west, magnitude $13\ \text{m/s}$, at $\tan^{-1}\tfrac{5.0}{12}=22.6^\circ$ to the vertical.

Answer (a): the rain appears to come from the east (ahead of her), so she tilts the umbrella forward, towards the east.

Exam tip (b) gets the direction backwards; (c) measures from the horizontal. The umbrella always points into the apparent rain.

🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. Why do elite shot-putters release the shot at about $37^\circ$, not at the "textbook" $45^\circ$?

Why it matters. Sports biomechanists film athletes at high speed and fit projectile models to their throws to advise coaches. Two effects, both missing from Result 3.9, pull the best angle below $45^\circ$: the shot is released about $2\ \text{m}$ above the ground, and the human body can push the shot faster at lower angles. Similar questions arise for long jump, javelin, basketball shots and even for the angle of irrigation sprinklers.

What to do.

Show from Result 3.8 that a projectile released at height $h$ with speed $u$ at angle $\theta$ lands at $R=\dfrac{u\cos\theta}{g}\left(u\sin\theta+\sqrt{u^2\sin^2\theta+2gh}\right)$.
In a spreadsheet, take $u=13.5\ \text{m/s}$ and $h=2.1\ \text{m}$ and compute $R$ for $\theta=20^\circ$ to $50^\circ$ in steps of $1^\circ$. Which angle is best? Compare with $h=0$.
Now model the athlete: release speed falls as the angle rises, $u=14.5-0.04(\theta-20^\circ)$ m/s. Find the new best angle.
Use Interactive 2 (with $h=0$) to see how the range curve flattens near its maximum — why does that make a few degrees "cheap"?

Tools: Spreadsheet; optionally a slow-motion phone video of a ball throw with a metre stick in view.

Hand in: A graph of $R$ against $\theta$ for the three models, the best angle in each, and a paragraph advising a young athlete.

Going further. Measure the release angle and speed of your own throw from a video (count frames) and compare the predicted range with the measured one. What causes the difference?

Instructor note: with $u=13.5\ \text{m/s}$, $h=0$ gives $45^\circ$ and $18.6\ \text{m}$; $h=2.1\ \text{m}$ gives a best angle of about $42^\circ$ ($20.6\ \text{m}$, versus $20.5\ \text{m}$ at $45^\circ$); adding the speed–angle trade-off moves the optimum to about $37^\circ$ ($21.2\ \text{m}$), close to what is measured for elite athletes. The flat top of $\sin2\theta$ means small angle errors cost little range — speed matters far more.

🧪 Self-check quiz

0/5 correct

Q1. For a projectile (no air resistance), which quantity stays constant throughout the flight?

Q2. A ball is thrown at $20\ \text{m/s}$ at $45^\circ$ on level ground. Taking $g=10\ \text{m/s}^2$, its range is

Q3. A projectile launched at $25^\circ$ has the same range as one launched with the same speed at

Q4. From the same height, ball P is dropped and ball Q is thrown horizontally at the same instant. Neglecting air resistance,

Q5. The velocity of a particle moving along a curved path is always

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