Projectile

Motion & Kinematics - OCR A-Level Physics

Projectile

Projectile
An object launched into the air that moves freely under gravity alone (ignoring air resistance). Its trajectory is a parabola.

Independence of components

The horizontal and vertical components of motion are independent. Vertically: $a = g$ downward, SUVAT applies. Horizontally: no acceleration, so $v_x$ is constant and $s_x = v_x t$. The same $t$ links the two motions.

Diagram pending

Parabolic trajectory of a projectile launched at angle $\theta$ above the horizontal at speed $u$. At launch, arrows show $u_x = u\cos\theta$ (horizontal) and $u_y = u\sin\theta$ (vertical). At highest point, only horizontal velocity arrow shown. At landing, both components shown again. Range $R$ marked on the ground. Maximum height $H$ marked vertically.

Will be replaced with a GeoGebra SVG in stream 2.

v_{x} = u\cos\theta

v_{y} = u\sin\theta

Case 1: Horizontal launch (e.g. ball off a cliff)

Initial vertical velocity $u_y = 0$. Time of flight set by the fall height $h$: $t = \sqrt{2h/g}$. Range: $R = v_x t$. Example: ball off a $45 \text{ m}$ cliff at $15 \text{ m s}^{-1}$ lands $45.4 \text{ m}$ away after $3.03 \text{ s}$.

Case 2: Angled launch from level ground

Resolve $u$: $u_x = u\cos\theta$, $u_y = u\sin\theta$. Time to highest point: $t_\text{up} = u\sin\theta / g$. Total time of flight (lands at same height): $T = 2 u\sin\theta / g$. Maximum height: $H = (u\sin\theta)^2 / (2g)$. Range on level ground: $R = u^2 \sin(2\theta) / g$.

Case 3: Angled launch from a height

Not symmetric. Solve the vertical equation $-h = u\sin\theta \cdot t - \tfrac{1}{2}gt^2$ as a quadratic in $t$ (taking up as positive). Use the positive root for $t$, then $R = u\cos\theta \cdot t$.

At the highest point of any trajectory, $v_y = 0$ but $v_x = u\cos\theta$ is unchanged. The object is still moving horizontally.
On level ground, the trajectory is symmetric: time up = time down, and launch speed equals landing speed.
Magnitude of velocity at landing: $v = \sqrt{v_x^2 + v_y^2}$. Direction: $\tan\phi = v_y / v_x$ below the horizontal.
For maximum range on level ground, launch at $\theta = 45^\circ$.

Common Mistake MEDIUM

Students often: A common slip: using the full launch speed in both horizontal and vertical SUVAT equations instead of resolving into components.
Instead: Resolve the initial velocity into horizontal (v cos theta) and vertical (v sin theta) components first. For a horizontal launch, the initial vertical velocity is zero.

Worked Example [4 marks]

A ball is kicked horizontally from the top of a cliff at 15 $m s^{-1}$. The cliff is 45 m high. Calculate the time to reach the ground and the horizontal distance travelled. Use g = 9.81 $m s^{-2}$.

Show Solution

Vertical

$s = 45 \text{ m}, u = 0 (horizontal launch), a = 9.81 \text{ m s}^{-2}$

[1]

$s = ut + \frac{1}{2}at^{2} \Rightarrow 45 = \frac{1}{2}(9.81)t^{2} \Rightarrow t = \sqrt{\frac{90}{9.81}} = 3.03 \text{ s}$

[1]

Horizontal

$no accelerationThe rate of change of velocity. A vector quantity. Measured in m s⁻²., so range = v_x \times t = 15 \times 3.03 = 45.4 \text{ m}$

[1]

Time to reach ground = 3.0 s; horizontal $distance = 45 m (2 s.f.)$

[1]

Answer

Time = 3.0 s; horizontal distance = 45 m

Examiner Tips and Tricks

Set up two columns of working: one for horizontal motion, one for vertical. List $u, v, a, s, t$ separately in each column. The shared variable is $t$.
Use the vertical motion to find $t$ first, then use that $t$ in the horizontal equation $s_x = v_x t$.
For "find the velocity on landing", combine the components with Pythagoras and state the angle below the horizontal.
OCR rounds answers in the final step. Keep extra significant figures in working.