Projectiles: horizontal launch, time of flight, height and range: WASSCE Physics

A footballer's long pass, water from a hose, a stone from a catapult: all follow the same curved path. The secret is to stop looking at the curve and look at two separate, simple motions happening at once.

Learning objectives

By the end of this lesson you should be able to:

define a projectile and describe its path
explain why horizontal and vertical motions are treated independently
solve problems on bodies projected horizontally from a height
calculate time of flight, maximum height and range for a body projected at an angle

What is a projectile?

Definition

A projectile is a body thrown into the air that then moves under the action of gravity alone (air resistance neglected). Its path is called its trajectory; it is a parabola.

Two independent motions

Parabolic path of a projectile: horizontal velocity arrows stay the same length along the path; vertical velocity arrows shrink to zero at the top and grow downwards after it

Figure 1. The horizontal velocity never changes; the vertical velocity changes by $g$ every second. WASSCE Tutor diagram.

	Horizontal motion	Vertical motion
Force acting	none (no air resistance)	weight, downwards
Acceleration	0	$g$ downwards
Velocity	constant	changes by $g$ each second
Equation	$x = u_x t$	$y = u_y t - \tfrac{1}{2}gt^2$

The two motions share only one thing: time. Find the time from one motion and use it in the other.

In Sierra Leone

Drop one orange from your hand and, at the same moment, flick another horizontally off the table edge at the same height. They hit the floor together. The horizontal flick does not delay the fall, because vertical and horizontal motions are independent.

Case 1: projected horizontally from a height

Initial velocity $u$ is horizontal, so $u_x = u$ and $u_y = 0$ .

Time to fall height $h$ : $h = \tfrac{1}{2}gt^2 \Rightarrow t = \sqrt{\dfrac{2h}{g}}$
Horizontal distance: $x = ut$
Velocity on landing: horizontal $u$ , vertical $gt$ ; speed $= \sqrt{u^2 + (gt)^2}$

Worked example

A stone is thrown horizontally at 15 m s⁻¹ from the top of a cliff 45 m high. Find (a) the time to reach the ground, (b) how far from the foot of the cliff it lands, (c) its speed on impact. ( $g = 10$ m s⁻²)

(a) $45 = \tfrac{1}{2}(10)t^2 \Rightarrow t^2 = 9 \Rightarrow t = 3$ s.

(b) $x = 15 \times 3 = 45$ m.

v = \sqrt{15^2 + 30^2} = \sqrt{1125} = 33.5\ \text{m s}^{-1}

\begin{aligned}v &= \sqrt{15^2 + 30^2} \\ &= \sqrt{1125} \\ &= 33.5\ \text{m s}^{-1}\end{aligned}

at $\tan^{-1}(30/15) = 63.4°$ below the horizontal.

Case 2: projected at an angle θ from the ground

Resolve the initial velocity:

u_x = u\cos\theta \qquad u_y = u\sin\theta

Quantity	Formula	Reason
Time to the highest point	$t = \dfrac{u\sin\theta}{g}$	vertical velocity becomes zero
Maximum height	$H = \dfrac{u^2\sin^2\theta}{2g}$	$0 = (u\sin\theta)^2 - 2gH$
Time of flight	$T = \dfrac{2u\sin\theta}{g}$	back to launch height
Horizontal range	$R = \dfrac{u^2\sin 2\theta}{g}$	$R = u\cos\theta \times T$

The range formula comes from $R = u\cos\theta \times \dfrac{2u\sin\theta}{g} = \dfrac{u^2(2\sin\theta\cos\theta)}{g}$ and $2\sin\theta\cos\theta = \sin 2\theta$ .

Worked example

A ball is kicked from the ground at 20 m s⁻¹ at 30° to the horizontal. Find (a) the time of flight, (b) the maximum height, (c) the range. ( $g = 10$ m s⁻²)

$u_x = 20\cos 30° = 17.32$ m s⁻¹, $u_y = 20\sin 30° = 10$ m s⁻¹.

(a) $T = \dfrac{2 \times 10}{10} = 2$ s.

(b) $H = \dfrac{10^2}{2 \times 10} = 5$ m.

Maximum range

$\sin 2\theta$ is largest (equal to 1) when $2\theta = 90°$ , that is θ = 45°:

R_{max} = \dfrac{u^2}{g}

Angles that add up to 90° (such as 30° and 60°) give the same range, because $\sin 60° = \sin 120°$ . The steeper throw goes higher and stays in the air longer.

Common mistake

At the highest point the projectile is not at rest. Its vertical velocity is zero, but its horizontal velocity $u\cos\theta$ is unchanged. Its speed at the top is $u\cos\theta$ , its smallest speed in the whole flight.

Key points

A projectile moves under gravity alone; its path is a parabola.
Horizontal: constant velocity $u\cos\theta$ . Vertical: acceleration $g$ downwards. Time links them.
Horizontal launch from height $h$ : $t = \sqrt{2h/g}$ , range $= ut$ .
At angle θ: $T = \dfrac{2u\sin\theta}{g}$ , $H = \dfrac{u^2\sin^2\theta}{2g}$ , $R = \dfrac{u^2\sin 2\theta}{g}$ ; maximum range at 45°.

WASSCE examination focus

WAEC tests this

Objective items ask for the angle of maximum range, the velocity at the highest point, or the time of flight. Theory items give a horizontal launch from a cliff or table, or a launch at an angle, and ask for time, range and impact velocity. Resolve the velocity first and keep horizontal and vertical working in separate columns.

Quick check

What is the vertical acceleration of a projectile at the top of its path?
A ball rolls off a table 1.25 m high at 4 m s⁻¹. How far from the table does it land? ( $g = 10$ m s⁻²)
Which angle of projection gives the same range as 25°?

Answers: 1. $g$ (10 m s⁻²) downwards. 2. $t = \sqrt{2(1.25)/10} = 0.5$ s; distance $= 4 \times 0.5 = 2$ m. 3. 65°.

Projectiles: horizontal launch, time of flight, height and range