Coaching Centre
Projectile motion
What a projectile is, the assumptions of the projectile model, resolving the launch velocity into components, why horizontal and vertical motion are independent, and solving horizontal-launch and angled-launch problems with the HSC equations.
Drop one coin off a desk and flick a second one sideways at the same moment. You will hear both hit the floor together. The flicked coin travels further, but it doesn't stay up any longer. That single observation is the whole key to projectile motion.
What counts as a projectile
A projectile is any object that, once launched, moves under the influence of gravity alone. A kicked football, a netball shot, water from a hose and a stone flicked off a cliff are all projectiles. A rocket with its engine running is not, and neither is a paper plane, because other forces matter.
The HSC model makes two assumptions:
- the acceleration is constant: $9.8\ \text{m s}^{-2}$ straight down, the same everywhere along the path;
- there is no air resistance, so gravity is the only force.
With those assumptions every projectile follows a parabola. They work well for dense, slow objects such as a shot put or a steel ball, and less well for light, fast ones such as a shuttlecock or a ping-pong ball. You'll see the effect of drag on page 5B.
Two motions, completely independent
Gravity acts straight down. It has no horizontal component, so it can't change horizontal velocity. That means a projectile's motion splits cleanly in two:
| Horizontal (x) | Vertical (y) | |
|---|---|---|
| Acceleration | zero | $a_y = -9.8\ \text{m s}^{-2}$ (taking up as positive) |
| Velocity | constant: $v_x = u_x$ | changes by 9.8 m/s every second: $v_y = u_y + a_y t$ |
| Displacement | $\Delta x = u_x t$ | $\Delta y = u_y t + \tfrac{1}{2}a_y t^2$ |
| Other | $v_x^2 = u_x^2$ | $v_y^2 = u_y^2 + 2a_y \Delta y$ |
These are the equations printed on the HSC formulae sheet. The only quantity the two columns share is time. Almost every projectile problem comes down to finding $t$ from one column and using it in the other.
Resolving the launch velocity
If an object is launched at speed $u$ at angle $\theta$ above the horizontal, split $u$ into components before you do anything else:
$u_x = u\cos\theta \qquad u_y = u\sin\theta$
At the highest point the vertical velocity is zero, but the horizontal velocity is still $u_x$. The projectile is never at rest unless it was launched straight up. At any moment its actual velocity is the vector sum of the components, so its speed is $v = \sqrt{v_x^2 + v_y^2}$ and its direction is $\tan^{-1}(v_y / v_x)$ from the horizontal.
A steel ball rolls off a bench 1.25 m high at 3.0 m/s. Find (a) how long it takes to land, (b) how far from the bench it lands and (c) its velocity on impact.
A footballer kicks a ball from the ground at 25 m/s, 40° above the horizontal. Ignoring air resistance, find the maximum height, the time of flight and the range.
- Set 20 m/s, 0° and a launch height of 20 m, with the second ball switched on. Launch. Which ball lands first? Check the time against $t = \sqrt{2h/g}$.
- Set the launch height to 0 and the speed to 20 m/s. Launch at 30°, then 60°. Compare the ranges and the times of flight.
- Try 45°. Is the range bigger or smaller than at 30°?
- Watch the blue arrow during a flight. What happens to it? What happens to the green arrow at the top of the path?
- Turn on air resistance and launch at 45° again. What happens to the range, and why is the way down steeper than the way up?
A method that always works
- Draw a sketch. Choose up as positive and mark the launch point as the origin.
- Resolve the initial velocity into $u_x$ and $u_y$.
- List the known values for each direction separately. Vertically, $a_y = -9.8\ \text{m s}^{-2}$; horizontally, $a_x = 0$.
- Find the time from whichever direction has enough information (usually vertical).
- Use the time in the other direction. Combine components at the end if you need a velocity.
Test yourself
A projectile is at the highest point of its path. Which statement is correct (no air resistance)?
Only the vertical component of velocity is zero; gravity still acts, so the acceleration is still g downward.
A ball is thrown horizontally at 8.0 m/s from a height of 4.9 m. How far does it travel horizontally before landing?
$t = \sqrt{2 \times 4.9/9.8} = 1.0$ s, so $\Delta x = 8.0 \times 1.0 = 8.0$ m.
A projectile is launched at 30 m/s at 30° above the horizontal. What is its velocity 1.0 s later?
$v_x = 30\cos30° = 26.0$; $v_y = 15 - 9.8 = 5.2$. So $v = \sqrt{26.0^2 + 5.2^2} = 26.5$ m/s at $\tan^{-1}(5.2/26.0) = 11.3°$ above the horizontal.
Two identical balls leave a table at the same instant: A rolls off at 2 m/s, B at 4 m/s. Which is correct?
The fall time depends only on the height. Range = $u_x t$, so doubling $u_x$ doubles the range.
Label the projectile's path.
Sort these quantities for a projectile with no air resistance.
A projectile is an object moving under the influence of alone. Its path is a . The horizontal component of velocity is because there is no horizontal force. At the highest point the component of velocity is zero. The only quantity shared by the horizontal and vertical motions is .
A stone is thrown horizontally at 12 m/s from the top of a cliff 45 m high. Ignoring air resistance, calculate the time it takes to reach the water below, how far from the base of the cliff it lands, and its velocity as it hits the water.
Take down as negative. Vertically, $u_y = 0$, $a_y = -9.8\ \text{m s}^{-2}$ and $\Delta y = -45$ m.
$-45 = \tfrac12(-9.8)t^2$, so $t^2 = 9.18$ and $t = 3.03$ s.
Horizontally, $\Delta x = u_x t = 12 \times 3.03 = 36.4$ m from the base of the cliff.
On impact, $v_y = 0 + (-9.8)(3.03) = -29.7\ \text{m s}^{-1}$ and $v_x = 12\ \text{m s}^{-1}$. The speed is $\sqrt{12^2 + 29.7^2} = 32.0\ \text{m s}^{-1}$, at $\tan^{-1}(29.7/12) = 68°$ below the horizontal.