Projectile motion

Physics · Senior school physics · about 10 min

A kicked ball, a thrown stone or a jet of water all follow the same curved path. The trick to every projectile problem is to treat the horizontal and vertical motion separately: they share only the time.

The key idea

Launched at speed $v_0$ and angle $\theta$ , ignoring air resistance:

v_x = v_0\cos\theta \;\;(\text{constant}) \qquad v_y = v_0\sin\theta - g t

On level ground: time of flight $t = \dfrac{2 v_0 \sin\theta}{g}$ , maximum height $H = \dfrac{(v_0\sin\theta)^2}{2g}$ , range $R = v_x\, t$ .

Two motions, one clock

Horizontally there's no force (ignoring air resistance), so the speed $v_x$ never changes: $x = v_x t$ .
Vertically gravity pulls down at $g$ , so it's motion with constant acceleration: $y = v_0\sin\theta\, t - \tfrac{1}{2} g t^2$ .

Find the time from the vertical motion (when does it land?), then use that time in the horizontal motion.

A smooth parabola from a launch point on level ground on the left, up to a peak and back down to the ground on the right. At three points, draw velocity parts as arrows: on the rising part, a dark blue arrow to the right and a dark red arrow up; at the peak, the same dark blue arrow to the right and no vertical arrow; on the falling part, the same dark blue arrow to the right and a dark red arrow down. All the dark blue arrows are the same length. Mark the launch speed and launch angle at the start. — The horizontal velocity stays the same while the vertical velocity changes.

The best angle

On level ground, the range is $R = \dfrac{v_0^2 \sin 2\theta}{g}$ , which is largest when $\sin 2\theta = 1$ , at $\theta = 45^\circ$ . Angles that add to $90^\circ$ (like $30^\circ$ and $60^\circ$ ) give the same range, but the steeper one goes higher and stays up longer.

Three smooth parabolas drawn on the same level ground, all starting from the same point on the left at the same speed. The 30 degree path and the 60 degree path land at the same point. The 45 degree path lands further right. Range in proportion: the 45 degree path is 1.15 times the other two. The 60 degree path is the highest. Draw the 45 degree path in dark blue and the other two in grey. — At 45 degrees the range is greatest. 30 and 60 degrees land at the same place.

Worked example: a ball kicked at 30°

A ball is kicked from level ground at $20$ m/s, $30^\circ$ above horizontal. Ignoring air resistance and taking $g = 9.81$ m/s $^2$ , find its time of flight, maximum height and range.

A smooth parabola from a launch point on level ground on the left to a landing point on the right. The range is 6.9 times the height. A dark blue arrow at the launch point, tangent to the path, with the 30 degree angle marked between it and the ground. A thin dashed vertical line from the peak to the ground. A dimension along the ground for the range. — The kick in this example.

Show the solution

Components: $v_x = 20\cos 30^\circ = 17.32$ m/s, $v_y = 20\sin 30^\circ = 10$ m/s

Time of flight:

t = \frac{2 \times 10}{9.81} = 2.04 \text{ s}

Maximum height:

H = \frac{10^2}{2 \times 9.81} = 5.10 \text{ m}

Range:

R = 17.32 \times 2.04 = 35.3 \text{ m}

Common mistakes

Letting gravity slow down the horizontal speed

Gravity only acts vertically. Without air resistance, $v_x$ is the same the whole flight.

Using the launch speed in both directions

Split it first: $v_0\cos\theta$ horizontally and $v_0\sin\theta$ vertically.

Thinking the speed is zero at the top

Only the vertical velocity is zero at the top; the ball still moves sideways at $v_x$ .

Check your understanding

1. At the highest point of its flight, a projectile's vertical velocity is...

2. Which launch angle gives the greatest range on level ground (no air resistance)?

3. A ball is thrown horizontally at $5$ m/s from a cliff and lands $2$ s later. How far from the base does it land?

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