Projectile and Uniform Circular Motion

Lessons

Senior 4 · Physics

Projectile and Uniform Circular Motion

5 lessons

Lesson 1 of 5

Projectiles and Horizontal Launch

25 min

Two coins, one sound

Lay a ruler on the edge of a desk with one coin balanced on its end and another coin just beside it on the desk. Strike the ruler so that it sweeps round: one coin is knocked straight off the desk horizontally, the other simply drops. Listen. However hard you strike, you hear a single click as both coins hit the floor together. The coin that flies off sideways travels much further, yet it takes no longer to fall. In this lesson you find out why, and use the answer to predict exactly where a thrown or launched object will land.

What a projectile is

Once a ball leaves a hand, a bullet leaves a gun or water leaves a hose, nothing pushes it forward any more. If air resistance is small, the only force on it is its weight, $m g$ , straight down. An object moving under its weight alone is a projectile, and the curved path it follows is its trajectory.

Figure 1.1 is drawn like a photograph taken with a flash every 0.10 s, of one ball dropped and another launched horizontally at the same moment. At every flash the two balls are at the same height, so they hit the floor together, as the coins did. Meanwhile the launched ball moves the same distance sideways in each interval.

Figure 1.1: A dropped ball and a ball launched horizontally at the same instant, imaged every 0.10 s (drawn to scale). Their vertical motions are identical; the launched ball also moves sideways at a steady speed.

Definition

A projectile is a body that, after being launched, moves under the action of gravity alone (air resistance neglected). Its path is its trajectory. The angle of projection $\theta$ is the angle between the launch velocity and the horizontal. The time of flight $T$ is the time from launch until it returns to the ground (or a stated level); the range $R$ is the horizontal distance it covers in that time; the maximum height $H$ is the greatest height it reaches above the launch point. Example: a football kicked from the ground at 15 m/s at $30^\circ$ above the horizontal is a projectile with $\theta = 30^\circ$ .

Two independent motions

The weight acts only vertically, so it changes only the vertical part of the velocity. This gives the key idea of the whole topic: the horizontal and vertical motions of a projectile are independent, linked only by the time $t$ , which is the same for both. Take the launch point as the origin, $x$ horizontal, $y$ vertical with up positive, and write the launch velocity as components $u_{x}$ and $u_{y}$ .

	Horizontal	Vertical (up positive)
acceleration	$a_{x} = 0$	$a_{y} = - g$
velocity	$v_{x} = u_{x}$ (constant)	$v_{y} = u_{y} - g t$
displacement	$x = u_{x} t$	$y = u_y t - \tfrac{1}{2} g t^{2}$
no time		$v_y^{2} = u_y^{2} - 2 g y$

Table 1.1: The equations of projectile motion; $g = 9.8\,\mathrm{m\,s^{-2}}$ near the Earth's surface.

These are the equations of uniformly accelerated motion you already know, applied separately in each direction. At any instant the speed is $v = \sqrt{v_x^{2} + v_y^{2}}$ and the velocity makes an angle $\varphi$ with the horizontal where $\tan\varphi = \frac{v_y}{v_x}$ .

Vertical projection

When a body is thrown straight up with speed $u$ , $u_{x} = 0$ and the motion is one-dimensional. At the top $v_{y} = 0$ , so

t_{\text{up}} = \frac{u}{g}, \qquad H = \frac{u^{2}}{2g}, \qquad T = \frac{2u}{g}

The motion is symmetric: the body takes as long to come down as to go up, and returns to the launch level with the same speed, now downwards.

Worked example 1: a ball thrown straight up

A ball is thrown vertically upwards at $14.7\,\mathrm{m\,s^{-1}}$ . Find the time to reach the top, the maximum height and the time of flight.

t_{\text{up}} = \frac{14.7}{9.8} = 1.50\,\mathrm{s}, \qquad H = \frac{14.7^{2}}{2 \times 9.8} = 11.0\,\mathrm{m}, \qquad T = 2 \times 1.50 = 3.00\,\mathrm{s}

Check with the displacement equation: at $t = 3.00\,\mathrm{s}$ , $y = 14.7 \times 3.00 - \tfrac{1}{2} \times 9.8 \times 3.00^{2} = 44.1 - 44.1 = 0$ , back at the hand.

Horizontal projection

A body launched horizontally with speed $u$ from a height $h$ has $u_{x} = u$ and $u_{y} = 0$ . It is convenient to measure the drop $y$ downwards from the launch point. Then

x = u t, \qquad y = \tfrac{1}{2} g t^{2}

It reaches the ground when $y = h$ , so the time of flight and range are

T = \sqrt{\frac{2h}{g}}, \qquad R = u T = u\sqrt{\frac{2h}{g}}

The time of flight depends only on the height, not on the launch speed: this is why the two coins land together. Eliminating $t = \frac{x}{u}$ gives the equation of the trajectory:

y = \frac{g}{2u^{2}}\,x^{2}

The drop grows as the square of the horizontal distance, so the path is a parabola. At time $t$ the velocity has components $v_{x} = u$ and $v_{y} = g t$ (downwards), so it turns steadily towards the vertical, as Figure 1.2 shows.

Figure 1.2: A stone thrown horizontally at 12 m/s from a 45 m cliff, drawn to scale. Velocity arrows at 1 s and 2 s: the horizontal component (blue) never changes; the vertical component (red) grows; their resultant (green) points along the path.

Worked example 2: a stone thrown from a cliff

A stone is thrown horizontally at $12\,\mathrm{m\,s^{-1}}$ from the top of a cliff 45 m high (Figure 1.2). Find (a) the time of flight, (b) the distance from the foot of the cliff where it lands, (c) its velocity on landing.

(a) $T = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 45}{9.8}} = 3.03\,\mathrm{s}$

(b) $R = uT = 12 \times 3.03 = 36.4\,\mathrm{m}$

v = \sqrt{12^{2} + 29.7^{2}} = 32.0\,\mathrm{m\,s^{-1}}, \qquad \tan\varphi = \frac{29.7}{12} \Rightarrow \varphi = 68.0^\circ \text{ below the horizontal}

Worked example 3: measuring a launch speed with a ruler

A marble rolls off a table 0.80 m high and lands 0.60 m from the foot of the table. Find the speed with which it left the table.

T = \sqrt{\frac{2 \times 0.80}{9.8}} = 0.404\,\mathrm{s}, \qquad u = \frac{R}{T} = \frac{0.60}{0.404} = 1.48\,\mathrm{m\,s^{-1}}

Only a metre rule is needed: the fall height fixes the time, and the range then gives the speed. This is a simple laboratory way to measure the speed of a ball leaving a ramp.

Warning

Common mistakes. (1) Using $g$ in the horizontal equation: the horizontal acceleration is zero. (2) Thinking that a faster horizontal launch makes a body stay up longer: the time of flight from a height $h$ is $\sqrt{\frac{2h}{g}}$ whatever the horizontal speed. (3) Adding the components as numbers: the landing speed is $\sqrt{v_x^{2} + v_y^{2}}$ , not $v_{x} + v_{y}$ . (4) Mixing sign conventions: choose up or down as positive at the start and keep it.

Tip

Split, solve, join. Split the launch velocity into horizontal and vertical parts; solve the vertical motion first, because it usually gives the time; then use that time in the horizontal motion; finally join the components with Pythagoras.

Shoot-n-Drop (Harvard Natural Sciences Lecture Demonstrations). An apparatus shoots one ball horizontally and drops another from the same height at the same moment. Watch, and listen, for both balls landing together, exactly as in Figure 1.1.

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Practice

A stone dropped from a bridge hits the water 2.4 s later. How high is the bridge above the water?
A ball is thrown vertically upwards at $20\,\mathrm{m\,s^{-1}}$ . Find its maximum height and the time it takes to return to the hand.
A ball rolls off a table 1.25 m high at $2.0\,\mathrm{m\,s^{-1}}$ . How far from the foot of the table does it land?
For the ball in question 3, find the speed and direction of its velocity just before it lands.
A dart is thrown horizontally at $15\,\mathrm{m\,s^{-1}}$ , aimed at the centre of a board 3.0 m away. How far below the centre does it hit?
An aircraft flying horizontally at $60\,\mathrm{m\,s^{-1}}$ at a height of 500 m drops a box of supplies. Ignoring air resistance, how far before the target must it release the box? Where is the aircraft when the box lands?
Show that the trajectory of a horizontally launched projectile is $y = \frac{g x^{2}}{2u^{2}}$ . If the launch speed is doubled, by what factor does the drop over a given horizontal distance change?
A student says: "A bullet fired horizontally stays in the air longer than one dropped from the same height, because it is moving so fast that gravity has less time to act on it." Evaluate this claim.

Show answersHide answers

$h = \tfrac{1}{2} \times 9.8 \times 2.4^{2} = 28\,\mathrm{m}$ .
$H = \frac{20^{2}}{2 \times 9.8} = 20.4\,\mathrm{m}$ ; $T = \frac{2 \times 20}{9.8} = 4.08\,\mathrm{s}$ .
$T = \sqrt{\frac{2 \times 1.25}{9.8}} = 0.505\,\mathrm{s}$ ; $R = 2.0 \times 0.505 = 1.01\,\mathrm{m}$ .
$v_y = 9.8 \times 0.505 = 4.95\,\mathrm{m\,s^{-1}}$ ; $v = \sqrt{2.0^{2} + 4.95^{2}} = 5.34\,\mathrm{m\,s^{-1}}$ at $\tan^{-1}\frac{4.95}{2.0} = 68^\circ$ below the horizontal.
$t = \frac{3.0}{15} = 0.20\,\mathrm{s}$ ; drop $= \tfrac{1}{2} \times 9.8 \times 0.20^{2} = 0.196\,\mathrm{m}$ , about 20 cm.
$T = \sqrt{\frac{2 \times 500}{9.8}} = 10.1\,\mathrm{s}$ ; release $60 \times 10.1 = 606\,\mathrm{m}$ before the target. The box keeps the aircraft's horizontal velocity, so the aircraft is directly above the box when it lands.
$x = ut \Rightarrow t = \frac{x}{u}$ ; $y = \tfrac{1}{2} g t^{2} = \frac{g x^{2}}{2u^{2}}$ . Doubling $u$ divides the drop by $2^{2} = 4$ .
Wrong. Gravity acts on both bullets in exactly the same way: both have $u_{y} = 0$ and $a_{y} = - g$ , so both fall the height in $\sqrt{\frac{2h}{g}}$ . The fast bullet simply travels much further sideways in that same time. (Air resistance and the curvature of the Earth make tiny differences that are negligible here.)

Quick check

A ball launched horizontally from a table takes longer to reach the floor than a ball dropped from the same table at the same moment.

Key points

A projectile moves under its weight alone; its horizontal and vertical motions are independent and share the same time.
Horizontally: $a_{x} = 0$ , $x = u_{x} t$ . Vertically: $a_{y} = - g$ , $v_{y} = u_{y} - g t$ , $y = u_y t - \tfrac{1}{2} g t^{2}$ .
Thrown straight up: $H = \frac{u^{2}}{2g}$ , $T = \frac{2u}{g}$ .
Launched horizontally from height $h$ : $T = \sqrt{\frac{2h}{g}}$ , $R = u\sqrt{\frac{2h}{g}}$ ; the path $y = \frac{g x^{2}}{2u^{2}}$ is a parabola.

Sources and credits

Photo: Bouncing ball strobe edit by MichaelMaggs, edited by Richard Bartz, Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: Shoot-n-Drop, Harvard Natural Sciences Lecture Demonstrations, embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, diagrams, table, worked examples and practice exercises. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Two coins, one sound

2What a projectile is

3Two independent motions

4Vertical projection

5Horizontal projection

Two coins, one sound

What a projectile is

Two independent motions

Vertical projection

Horizontal projection