Motion in a Plane • Topic 2 of 3

Projectile Motion

Throw a stone at an angle and watch it: it rises, curves over and falls in a graceful arch. Any object thrown into the air that then moves only under gravity (we ignore air resistance) is called a projectile, and its curved path is its trajectory. The whole subject of projectile motion rests on one brilliant idea — the horizontal and vertical motions are completely independent of each other and can be analysed separately.

Suppose the projectile is launched with speed $u$ at an angle $\theta$ above the horizontal. We split this into a horizontal part $u_x=u\cos\theta$ and a vertical part $u_y=u\sin\theta$. Horizontally there is no acceleration, so the horizontal velocity stays constant: $x=u\cos\theta\cdot t$. Vertically the only force is gravity, giving a constant downward acceleration $g$, so $y=u\sin\theta\cdot t-\tfrac{1}{2}gt^2$. Eliminating $t$ between these gives the equation of the path: $y=x\tan\theta-\dfrac{g\,x^2}{2u^2\cos^2\theta}$, which is the equation of a parabola. So every projectile traces a parabola.

Three quantities describe the flight. The time of flight $T$ is how long the projectile stays in the air; since the vertical velocity must reverse and the projectile returns to launch height, $T=\dfrac{2u\sin\theta}{g}$. The maximum height $H$ is reached when the vertical velocity becomes zero: $H=\dfrac{u^2\sin^2\theta}{2g}$. The horizontal range $R$ is the horizontal distance covered in the total time of flight: $R=u\cos\theta\cdot T=\dfrac{u^2\sin 2\theta}{g}$.

A neat consequence: range is maximum when $\sin 2\theta=1$, i.e. $\theta=45^\circ$, giving $R_{max}=\dfrac{u^2}{g}$. Also, two angles that add up to $90^\circ$ (like $30^\circ$ and $60^\circ$) give the same range. A special case is the horizontal projectile — an object thrown horizontally with speed $u$ from a height $h$. Here $u_y=0$ initially, so the time to fall is $t=\sqrt{\dfrac{2h}{g}}$ and the horizontal range is $R=u\sqrt{\dfrac{2h}{g}}$, while its path is still half a parabola.

Path: a parabola, $y=x\tan\theta-\dfrac{g\,x^2}{2u^2\cos^2\theta}$.
Time of flight: $T=\dfrac{2u\sin\theta}{g}$.
Maximum height: $H=\dfrac{u^2\sin^2\theta}{2g}$.
Range: $R=\dfrac{u^2\sin 2\theta}{g}$; maximum at $\theta=45^\circ$ where $R_{max}=\dfrac{u^2}{g}$.
Horizontal projectile from height $h$: $t=\sqrt{\dfrac{2h}{g}}$, $R=u\sqrt{\dfrac{2h}{g}}$.

Solved Examples

Worked Example

A ball is projected at $20\,\text{m/s}$ at an angle of $30^\circ$ to the horizontal. Find its time of flight. (Take $g=10\,\text{m/s}^2$.)

Solution

$T=\dfrac{2u\sin\theta}{g}$ with $u=20$, $\theta=30^\circ$, $\sin 30^\circ=\tfrac{1}{2}$.
$T=\dfrac{2\times 20\times\tfrac{1}{2}}{10}=\dfrac{20}{10}$.
$=2\,\text{s}$.

Answer: Time of flight $=2\,\text{s}$.

Worked Example

For the same ball ($u=20\,\text{m/s}$, $\theta=30^\circ$, $g=10\,\text{m/s}^2$), find the maximum height reached.

Solution

$H=\dfrac{u^2\sin^2\theta}{2g}$ with $\sin 30^\circ=\tfrac{1}{2}$, so $\sin^2 30^\circ=\tfrac{1}{4}$.
$H=\dfrac{20^2\times\tfrac{1}{4}}{2\times 10}=\dfrac{400\times\tfrac{1}{4}}{20}$.
$=\dfrac{100}{20}=5\,\text{m}$.

Answer: Maximum height $=5\,\text{m}$.

Worked Example

A projectile is fired at $u=40\,\text{m/s}$ at $45^\circ$. Find its horizontal range. (Take $g=10\,\text{m/s}^2$.)

Solution

$R=\dfrac{u^2\sin 2\theta}{g}$ with $2\theta=90^\circ$, so $\sin 90^\circ=1$.
$R=\dfrac{40^2\times 1}{10}=\dfrac{1600}{10}$.
$=160\,\text{m}$ (this is also the maximum possible range for this speed).

Answer: Range $=160\,\text{m}$.

Worked Example

A stone is thrown at $u=10\,\text{m/s}$ at $60^\circ$. Find its range. (Take $g=10\,\text{m/s}^2$, $\sin 120^\circ=\tfrac{\sqrt{3}}{2}$.)

Solution

$R=\dfrac{u^2\sin 2\theta}{g}$ with $2\theta=120^\circ$.
$R=\dfrac{10^2\times\tfrac{\sqrt{3}}{2}}{10}=\dfrac{100\times 0.866}{10}$.
$=\dfrac{86.6}{10}=8.66\,\text{m}$.

Answer: Range $\approx 8.66\,\text{m}$.

Worked Example

An object is thrown horizontally at $u=15\,\text{m/s}$ from a cliff $20\,\text{m}$ high. Find the time to reach the ground. (Take $g=10\,\text{m/s}^2$.)

Solution

Vertical motion only: $t=\sqrt{\dfrac{2h}{g}}$ with $h=20\,\text{m}$.
$t=\sqrt{\dfrac{2\times 20}{10}}=\sqrt{4}$.
$=2\,\text{s}$ (independent of the horizontal speed).

Answer: Time to land $=2\,\text{s}$.

Worked Example

For the horizontal projectile above ($u=15\,\text{m/s}$, $h=20\,\text{m}$, $t=2\,\text{s}$), find the horizontal distance covered before landing.

Solution

Horizontal velocity is constant, so $R=u\cdot t$.
$R=15\times 2$.
$=30\,\text{m}$.

Answer: Horizontal distance $=30\,\text{m}$.

Key Points

A projectile moves under gravity alone; its horizontal and vertical motions are independent.
The trajectory is a parabola: $y=x\tan\theta-\dfrac{g\,x^2}{2u^2\cos^2\theta}$.
Time of flight $T=\dfrac{2u\sin\theta}{g}$ and maximum height $H=\dfrac{u^2\sin^2\theta}{2g}$.
Range $R=\dfrac{u^2\sin 2\theta}{g}$ is maximum at $\theta=45^\circ$, where $R_{max}=\dfrac{u^2}{g}$.
For a horizontal projectile from height $h$: $t=\sqrt{\dfrac{2h}{g}}$ and $R=u\sqrt{\dfrac{2h}{g}}$.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.The path (trajectory) of a projectile is a:

Explanation: Eliminating time between horizontal and vertical equations gives y as a quadratic in x, i.e. a parabola.

Q2.The horizontal range of a projectile is maximum when the angle of projection is:

Explanation: R = u^2 sin(2 theta)/g is maximum when sin(2 theta) = 1, i.e. 2 theta = 90, so theta = 45 degrees.

Q3.During projectile motion, the horizontal component of velocity:

Explanation: There is no horizontal acceleration, so u cos theta stays constant throughout the flight.

Q4.At the highest point of its path, a projectile's vertical velocity is:

Explanation: At maximum height the upward motion stops momentarily, so the vertical velocity is zero (only the horizontal part remains).

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