Range and maximum height of a projectile — Physics

Before this

This leads to

Projectiles launched at an angle

Further: more speed, or more height?

In the last lecture, a ball launched horizontally at 12 m/s from a 20 m ledge landed about 24.2 m from the foot of the ledge. Suppose you want it to land further away. You can do one of two things: double the launch speed, or launch from a ledge twice as high. Which gives the longer distance? Decide before you read on.

Predict — double the speed or double the height?

The ball starts at 12 m/s from a 20 m ledge and lands 24.2 m out. Which change makes it land further away? Choose one, then read the explanation.

The two changes, worked out with g = 9.8 m/s²:

Double the speed (24 m/s from 20 m): the time in the air is still 2.02 s, so the ball lands $24 \times 2.02 \approx 48.5$ m out. The distance doubles.
Double the height (12 m/s from 40 m): the time becomes 2.86 s, only about 1.41 times as long, so the ball lands $12 \times 2.86 \approx 34.3$ m out.

Doubling the speed wins: 48.5 m against 34.3 m.

Most people expect the two changes to be equal. They are not, and the reason is in how falling works. This lecture finds the distance a projectile travels and the greatest height it reaches. It also explains why speed and height affect the distance so differently.

Range: the vertical list gives the time

The range of a projectile is the horizontal distance from where it is launched to where it lands. The method is the one from the last lecture: two SUVAT lists, linked by the time.

A horizontal launch from height H. The ball curves down and lands a distance R, the range, from the foot of the ledge (drawn to scale for u = 12 m/s and H = 20 m).

For a horizontal launch from a height $H$ , take downwards as positive in the vertical list. The vertical velocity starts at zero, and the acceleration is $g$ . So $s = u t + \frac{1}{2} a t^{2}$ becomes $H = \frac{1}{2} g t^{2}$ . Rearranging gives the time of flight:

t = \frac{2 H}{g}

The horizontal list has no acceleration, so the range is the launch speed times this time:

R = u \times t = u \frac{2 H}{g}

Range of a horizontal launch

Launched horizontally at speed $u$ from height $H$ , a projectile is in the air for $t = 2 H \div g$ , and its range is $R = u 2 H \div g$ .

You do not need to learn this formula. It is just the two lists written in one line. If you forget it, write the two lists again. That takes less than a minute, and it works for problems the formula does not cover.

Why height counts for less than speed

Now the hook can be answered. Look at what each quantity does in $R = u 2 H \div g$ .

The speed $u$ multiplies the range directly. It plays no part in the time, because the time comes from the vertical list. So twice the speed means twice the distance in the same time.

The height $H$ sits under a square root. This is because a falling object speeds up. It covers only 4.9 m in its first second, but 14.7 m in its second second. So to fall twice as far, it does not need twice as long. Doubling the height multiplies the time by $2 \approx 1.41$ , and the range by the same amount.

That is it. That is the whole idea. Speed works on the horizontal list, which is simple and steady. Height works through the vertical list, where the motion speeds up, so extra height gives diminishing returns. To multiply the range by 2 using height alone, you would need a ledge 4 times as high.

The simulation below lets you test this. Choose a ledge, choose a launch speed, and try to hit a target 30 m away.

Simulation — hit the target

Choose a ledge height and a launch speed. Press Launch, or step through the flight. Can you land within 0.5 m of the target at 30 m from each ledge?

Launch speed: 12 m/s

The speed needed to land 30 m out from each ledge ( $u = 30 \div t$ ):

Ledge height	Time in the air	Speed needed
5 m	1.01 s	29.7 m/s
20 m	2.02 s	14.8 m/s
45 m	3.03 s	9.9 m/s

Each ledge is 4 times or 9 times as high as the 5 m ledge, but the time is only 2 or 3 times as long. So the speed needed falls to only a half or a third.

Did you find the three speeds? From 5 m you need about 29.7 m/s, from 20 m about 14.8 m/s, and from 45 m about 9.9 m/s. The 45 m ledge is 9 times as high as the 5 m one, but it gives only 3 times as long in the air, because $9 = 3$ .

Working backwards

The two lists can be used in either order. If the range is known, the horizontal list gives the time, and the vertical list then gives the height. The worked examples below show both directions.

Maximum height

For a horizontal launch, the question "how high does it go?" has a simple answer. The ball starts with no upward velocity, so it only ever moves down. Its greatest height is the height it starts from.

The question becomes interesting when a ball is thrown sideways and upwards. Suppose a ball leaves your hand moving at 20 m/s sideways and 12 m/s upwards at the same moment. (The next lecture shows how these two parts come from the speed and angle of a throw.) The two lists are still separate. Take upwards as positive in the vertical list this time, because the ball starts by moving up.

The ball keeps rising while its vertical velocity is upwards. Gravity reduces that velocity by 9.8 m/s every second. At the top, the vertical velocity has fallen to zero. So the vertical list is: $u = + 12$ m/s, $v = 0$ , $a = - 9.8 m/s^{2}$ , $s = ?$

v^{2} = u^{2} + 2 a s \Rightarrow 0 = 144 - 19.6 s \Rightarrow s = \frac{144}{19.6} \approx 7.35 m

The time to reach the top comes from $v = u + a t$ : $0 = 12 - 9.8 t$ , so $t \approx 1.22$ s. In that time, the horizontal list moves the ball $20 \times 1.22 \approx 24.5$ m sideways.

Notice what was not used. The 20 m/s sideways speed played no part in the greatest height. Only the upward velocity decides how high a projectile goes.

A ball thrown at 20 m/s sideways and 12 m/s upwards. At the top, 7.35 m above the launch level, its vertical velocity is zero, but it is still moving sideways at 20 m/s. (The vertical scale is stretched to make the height easy to see.)

Stop and think

At the highest point of its path, is the ball's velocity zero?

Show the answer

No. Only its vertical velocity is zero there. Its horizontal velocity is still 20 m/s, because nothing acts sideways. So at the top, the ball is moving horizontally at 20 m/s. A ball thrown straight up does stop for an instant at the top, because it has no sideways motion. A projectile thrown sideways never stops.

Worked examples

In each example, state the positive direction for the vertical list first.

Worked Example 1 — A package off a shelf

A package slides off a shelf horizontally at 4 m/s. The shelf is 1.8 m above the floor. How far from the shelf does it land?

Vertical (down positive): $1.8 = 4.9 t^{2}$ , so $t^{2} \approx 0.367$ and $t \approx 0.606$ s.

Horizontal: $R = 4 \times 0.606 \approx 2.42$ m.

Check: 4 m/s for about 0.6 s is about 2.4 m.

Worked Example 2 — From the range to the height

A ball is launched horizontally at 18 m/s from the top of a tower. It lands 54 m from the foot of the tower. How tall is the tower?

Horizontal first, because the range and the speed are both known: $t = 54 \div 18 = 3$ s.

Vertical (down positive): $u = 0$ , $a = 9.8 m/s^{2}$ , $t = 3$ s. So $s = \frac{1}{2} \times 9.8 \times 3^{2} = 4.9 \times 9 = 44.1$ m.

The tower is 44.1 m tall. Check with the range formula: $18 \times 2 \times 44.1 \div 9.8 = 18 \times 3 = 54$ m.

Worked Example 3 — How fast to reach a cup?

A marble rolls off a table 0.8 m high. It must land in a cup on the floor, 2.0 m from the edge of the table. How fast must it leave the table?

Vertical (down positive): $0.8 = 4.9 t^{2}$ , so $t \approx 0.404$ s. This time is fixed by the table, whatever the speed.

Horizontal: $u = 2.0 \div 0.404 \approx 4.95$ m/s.

Check: $4.95 \times 0.404 \approx 2.0$ m.

Worked Example 4 — The top of an upward throw

A ball leaves a player's foot at 9 m/s sideways and 15 m/s upwards. Find its greatest height above the ground and how far it has travelled sideways at that moment.

Vertical (up positive): $u = + 15$ m/s, $v = 0$ at the top, $a = - 9.8 m/s^{2}$ . From $v^{2} = u^{2} + 2 a s$ : $0 = 225 - 19.6 s$ , so $s \approx 11.5$ m. From $v = u + a t$ : $t = 15 \div 9.8 \approx 1.53$ s.

Horizontal: $9 \times 1.53 \approx 13.8$ m.

The ball reaches about 11.5 m, about 13.8 m from where it was kicked. Check: the upward velocity is the same 15 m/s as the ball in the kinematics lectures, and that ball also rose 11.5 m. The sideways motion changes nothing about the height.

Three classic mistakes

Trap 1 — "Twice the height, twice the range"

The time of flight grows with the square root of the height, so doubling the height multiplies the range by only about 1.41. To double the range by height alone, the height must be 4 times as great.

Trap 2 — "The velocity is zero at the top"

Only the vertical velocity is zero at the highest point. The horizontal velocity is unchanged. If you set the whole velocity to zero, the horizontal list will give wrong answers.

Trap 3 — Using the sideways speed for the height

The greatest height depends only on the upward velocity. Putting the sideways speed into $v^{2} = u^{2} + 2 a s$ gives a height that is far too large. Keep each velocity in its own list.

Try it yourself

Work each one on paper first. Take g = 9.8 m/s² and ignore air resistance. Open a hint only when you are stuck.

A stone is thrown horizontally at 15 m/s from a cliff 30 m high. Find its time of flight and its range.

Hint 1
The vertical list gives the time: $30 = 4.9 t^{2}$ .

Hint 2
Then range = speed × time.

Full answer
$t = 30 \div 4.9 \approx 2.47$ s. Range $= 15 \times 2.47 \approx 37.1$ m. Check with the formula: $15 2 \times 30 \div 9.8 \approx 37.1$ m.
A ball rolled off a table lands 0.6 m from the table. Where would it land if it left the table (a) three times as fast, (b) at the same speed from a table four times as high?

Hint 1
Speed multiplies the range directly. Height acts through a square root.

Hint 2
$4 = 2$ .

Full answer
(a) Three times the speed, same time: $3 \times 0.6 = 1.8$ m. (b) Four times the height gives $4 = 2$ times the time, so $2 \times 0.6 = 1.2$ m. No numbers for the speed or height were needed: only how the range depends on each.
A ball is thrown at 6 m/s sideways and 9.8 m/s upwards. Find its greatest height above the launch point and the time it takes to get there. What is its velocity at that moment?

Hint 1
Up positive. At the top, the vertical velocity is zero.

Hint 2
$0 = 9. 8^{2} - 19.6 s$ , and $0 = 9.8 - 9.8 t$ .

Full answer
$s = 96.04 \div 19.6 = 4.9$ m, and $t = 1$ s. At the top, the velocity is 6 m/s horizontally: the vertical part is zero, and the horizontal part has not changed.
The ball in question 3 is thrown from the edge of a flat roof 25 m above the ground. How long is it in the air altogether, and how far from the building does it land?

Hint 1
Split the flight at the top. You already know the first part: 1 s up, to 4.9 m above the roof.

Hint 2
From the top, the ball falls from rest through $25 + 4.9 = 29.9$ m. Use $29.9 = 4.9 t^{2}$ .

Full answer
Up: 1 s. Down from the top (down positive, starting from zero vertical velocity): $t = 29.9 \div 4.9 \approx 2.47$ s. Total time $\approx 1 + 2.47 = 3.47$ s. Range $= 6 \times 3.47 \approx 20.8$ m. Check: the time down is longer than the time up, as it must be, because the ball falls further than it rose.

The bigger idea

The square root in $t = 2 H \div g$ is a signature of steady acceleration. Whenever something speeds up at a constant rate from rest, the distance grows with the square of the time, so the time grows only with the square root of the distance. You will see the same pattern wherever acceleration is constant.

The range and the greatest height are the two questions asked about every projectile. In the next lecture, a ball is launched at an angle, and both questions are answered in one complete analysis, using exactly the methods in this lecture.

Key concepts in this lesson

The range is the horizontal distance from launch to landing, found by getting the time from the vertical list and using it in the horizontal list.
For a horizontal launch from height H at speed u, the time of flight is √(2H ÷ g) and the range is u√(2H ÷ g).
Doubling the launch speed doubles the range, but doubling the height multiplies the range by only about 1.41.
A projectile's greatest height depends only on its upward velocity, and it is reached when the vertical velocity becomes zero.
At the highest point the vertical velocity is zero, but the horizontal velocity is unchanged, so the projectile is still moving.