Projectile launched at an angle — full analysis — Physics

Before this

Range and maximum height
Projectile motion — horizontal and vertical are independent
SOH CAH TOA · coming soon

This leads to

One speed, one angle, two questions

A football is kicked from the ground at 25 m/s, at 35° above the horizontal. How high does it rise, and how far away does it land? If air resistance is ignored, the answers are about 10.5 m and about 59.9 m. This lecture shows how to get them.

The last two lectures did almost all the work. They showed that a projectile's horizontal and vertical motions are independent, linked only by the time. They found the range and the greatest height when the two parts of the velocity were given separately. The last lecture promised to show where those two parts come from when you know only the speed and the angle. Here is that last piece.

Splitting a velocity into two parts

A velocity at an angle is really two velocities at once: one sideways and one upwards. Draw the launch velocity as an arrow. Then draw a right-angled triangle with that arrow as its longest side. The other two sides are the horizontal part and the vertical part.

A launch at 20 m/s and 30°, drawn to scale. The arrow is the hypotenuse of a right-angled triangle. Its horizontal side is the horizontal velocity, and its vertical side is the vertical velocity.

Right-angled triangles are exactly what sine and cosine are for. With a launch speed $u$ at an angle $θ$ above the horizontal, the side next to the angle is the horizontal part, and the side opposite the angle is the vertical part:

u_{x} = u cos θ u_{y} = u sin θ

Here $u_{x}$ means "the horizontal part of $u$ " and $u_{y}$ means "the vertical part of $u$ ". Splitting a velocity like this is called resolving it into components. For 20 m/s at 30°: $u_{x} = 20 cos 30° \approx 17.3$ m/s and $u_{y} = 20 sin 30° = 10$ m/s.

Now the promise from the last lecture. That lecture used a ball moving at 20 m/s sideways and 12 m/s upwards. Those two parts form a right-angled triangle, so the launch speed is $2 0^{2} + 1 2^{2} \approx 23.3$ m/s, and the angle is the one whose tangent is $12 \div 20$ , about 31°. So that ball was kicked at about 23.3 m/s at about 31°. Resolving works in both directions.

Check the split by sense. At 0°, the whole speed is horizontal: $cos 0° = 1$ and $sin 0° = 0$ . At 90°, the whole speed is vertical. In between, the speed is shared, and a steeper angle gives more of it to the vertical part.

The full analysis, in five steps

Once the velocity is resolved, the method is the one you already know: two SUVAT lists, linked by the time. For a projectile launched from the ground and landing on the same ground, take upwards as positive in the vertical list.

Resolve: $u_{x} = u cos θ$ and $u_{y} = u sin θ$ .
Time to the top: the vertical velocity falls to zero, so $0 = u_{y} - g t$ , and $t_{top} = u_{y} \div g$ .
Total time: the flight down mirrors the flight up, so the total time is $T = 2 t_{top}$ .
Greatest height: from $v^{2} = u^{2} + 2 a s$ with $v = 0$ : $H = u_{y}^{2} \div 2 g$ .
Range: the horizontal velocity is steady for the whole flight, so $R = u_{x} \times T$ .

Why is the flight symmetric? Going up, gravity takes away 9.8 m/s of vertical velocity every second until none is left. Coming down, it adds 9.8 m/s every second. Falling from the top back to the same level is the climb played backwards, so it takes the same time. This is true only when the ball lands at the height it started from.

Step through the five steps below for a ball launched at 20 m/s and 30°.

Step through — 20 m/s at 30°, the full analysis

Press Next to add one step. Read the reason before you go on.

Each step gives one line and the reason for it. Take g = 9.8 m/s² and upwards as positive.

Step 1. horizontal part = 20 cos 30° ≈ 17.3 m/s and vertical part = 20 sin 30° = 10 m/s: resolve the launch velocity into its two parts.
Step 2. t_top = 10 ÷ 9.8 ≈ 1.02 s: at the top, the vertical velocity has fallen from 10 m/s to zero, losing 9.8 m/s each second.
Step 3. T = 2 × 1.02 ≈ 2.04 s: the ball lands at the level it started from, so the way down takes as long as the way up.
Step 4. H = 10² ÷ (2 × 9.8) ≈ 5.10 m: from v² = u² + 2as with v = 0 and a = −9.8 m/s².
Step 5. R = 17.3 × 2.04 ≈ 35.3 m: the horizontal velocity is steady for the whole flight.

Result: the ball is in the air for about 2.04 s, rises about 5.10 m, and lands about 35.3 m away.

Stop and think

A student uses the whole launch speed, 20 m/s, as the vertical velocity in step 2. What answer for the greatest height would that give, and why is it wrong?

Show the answer

It would give $2 0^{2} \div 19.6 \approx 20.4$ m, four times too high. That is the height for a ball thrown straight up at 20 m/s. At 30°, only the vertical part, 10 m/s, fights gravity. The other part of the velocity is carrying the ball sideways, and sideways motion does nothing to lift it. Always resolve first.

Which angle goes furthest?

Keep the speed at 20 m/s and change the angle. A low angle gives a large horizontal part, but the ball is soon back on the ground. A steep angle keeps the ball in the air for a long time, but it hardly moves sideways. The range depends on both, so somewhere between there must be a best angle.

Write the range using the five steps: $R = u_{x} \times T = u cos θ \times \frac{2 u sin θ}{g}$ . So the range is proportional to $sin θ \times cos θ$ . Now notice something. $sin 60° = cos 30°$ , and $cos 60° = sin 30°$ , because the two angles of a right-angled triangle add to 90°. So $sin 30° cos 30°$ and $sin 60° cos 60°$ are the same product, just multiplied in a different order. The ranges at 30° and at 60° are equal.

That is it. That is the whole idea. Any two angles that add to 90° give the same range, one with a low, fast flight and one with a high, slow flight. The best angle is the one that pairs with itself: 45°. At 45°, the two parts of the velocity are equal, and the range is greatest. Explore it below.

Explore — the launch angle

The launch speed is 20 m/s. Change the angle. The solid path is your angle; the dashed path is the angle that adds up to 90° with it.

Angle: 30°

Launches at 20 m/s from level ground (g = 9.8 m/s²):

Angle	Time in the air	Greatest height	Range
15°	1.06 s	1.37 m	20.4 m
30°	2.04 s	5.10 m	35.3 m
45°	2.89 s	10.2 m	40.8 m
60°	3.53 s	15.3 m	35.3 m
75°	3.94 s	19.0 m	20.4 m

15° and 75° give the same range, and so do 30° and 60°. The greatest range, 40.8 m, is at 45°.

Did you notice the pairs? 15° and 75° both reach about 20.4 m, and 30° and 60° both reach about 35.3 m. The pairs land in the same place, but the steeper one stays in the air much longer and rises much higher.

One caution. The 45° rule assumes no air resistance and level ground. Real balls are slowed by the air, and many real launches start above the ground, like a throw from a player's hand. Both change the best angle. For a first model, though, 45° is the right answer, and the reasoning behind it is what matters.

When the landing is lower than the launch

Step 3, $T = 2 t_{top}$ , works only when the ball lands at the height it started from. If it is launched from a cliff top, it falls further on the way down than it rose on the way up, so the flight is not symmetric. The fix is the method from the last lecture's final question: split the flight at the top. Find the time up, then the time to fall from the top all the way to the ground, starting with zero vertical velocity. Add the two times, and use the total in the horizontal list.

Worked examples

In each example, take g = 9.8 m/s², ignore air resistance, and take upwards as positive.

Worked Example 1 — The free kick

A ball is kicked from the ground at 25 m/s at 35°. Find its greatest height and its range.

Resolve: $u_{x} = 25 cos 35° \approx 20.5$ m/s and $u_{y} = 25 sin 35° \approx 14.3$ m/s.

Height: $H = 14. 3^{2} \div 19.6 \approx 10.5$ m.

Time: $t_{top} = 14.3 \div 9.8 \approx 1.46$ s, so $T \approx 2.93$ s.

Range: $R = 20.5 \times 2.93 \approx 59.9$ m.

Check: the range should be a little less than the 45° maximum for this speed, $2 5^{2} \div 9.8 \approx 63.8$ m. It is.

Worked Example 2 — A javelin at 45°

A javelin is thrown at 15 m/s at 45° from ground level. Find its range, treating it as a projectile with no air resistance.

$u_{x} = u_{y} = 15 cos 45° \approx 10.6$ m/s. So $T = 2 \times 10.6 \div 9.8 \approx 2.16$ s, and $R \approx 10.6 \times 2.16 \approx 22.9$ m.

At 45°, $sin θ cos θ = \frac{1}{2}$ , so the range formula above becomes $R = u^{2} \div g = 225 \div 9.8 \approx 23.0$ m. The two answers agree, apart from rounding.

Worked Example 3 — The pair that lands together

A ball is launched at 20 m/s at 60°. Find its time of flight, greatest height and range, and compare them with the 30° launch in the step-through.

$u_{x} = 20 cos 60° = 10$ m/s, $u_{y} = 20 sin 60° \approx 17.3$ m/s. Then $T = 2 \times 17.3 \div 9.8 \approx 3.53$ s, $H = 17. 3^{2} \div 19.6 \approx 15.3$ m, and $R = 10 \times 3.53 \approx 35.3$ m.

The range matches the 30° launch, as predicted, but the ball is in the air about 1.7 times as long and rises 3 times as high. The two parts of the velocity have simply swapped.

Three classic mistakes

Trap 1 — Not resolving

Using the whole launch speed in the vertical or horizontal list is the most common error. Only $u sin θ$ is vertical and only $u cos θ$ is horizontal. Resolve first, every time.

Trap 2 — Swapping sine and cosine

When the angle is measured from the horizontal, the horizontal part uses cosine and the vertical part uses sine. If a question gives the angle from the vertical, they swap. Draw the triangle and check: a steeper launch must give a bigger vertical part.

Trap 3 — Doubling the time when the levels differ

$T = 2 t_{top}$ is true only when the launch and landing heights are equal. From a cliff or onto a roof, split the flight at the top and work out the two parts separately.

Try it yourself

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

A ball is launched from the ground at 20 m/s at 45°. Find its time of flight, greatest height and range.

Hint 1
$u_{x} = u_{y} = 20 cos 45° \approx 14.1$ m/s.

Hint 2
$t_{top} = 14.1 \div 9.8$ , then double it.

Full answer
$t_{top} \approx 1.44$ s, so $T \approx 2.89$ s. $H = 14. 1^{2} \div 19.6 \approx 10.2$ m. $R \approx 14.1 \times 2.89 \approx 40.8$ m. Check with $R = u^{2} \div g = 400 \div 9.8 \approx 40.8$ m.
Without calculating, which launch at 20 m/s has the greater range: 20° or 70°? Which stays in the air longer?

Hint 1
Do the two angles add to 90°?

Hint 2
Which angle gives the bigger vertical part?

Full answer
20° + 70° = 90°, so the ranges are equal. The 70° launch has the bigger vertical part, so it stays in the air longer and rises higher; the 20° launch makes up for its shorter time with a bigger horizontal part.
A ball leaves the ground moving at 9 m/s horizontally and 15 m/s vertically. Find its launch speed and angle.

Hint 1
The two parts are the sides of a right-angled triangle. The speed is the hypotenuse.

Hint 2
Use Pythagoras for the speed, and $tan θ = u_{y} \div u_{x}$ for the angle.

Full answer
Speed $= 9^{2} + 1 5^{2} = 306 \approx 17.5$ m/s. $tan θ = 15 \div 9 \approx 1.67$ , so $θ \approx 59°$ . (This is the kicked ball from Worked Example 4 of the last lecture.) Check: $17.5 cos 59° \approx 9.0$ and $17.5 sin 59° \approx 15.0$ .
The 20 m/s, 30° ball from the step-through is launched from the top of a cliff 15 m high instead of from the ground. How far from the foot of the cliff does it land?

Hint 1
The flight is not symmetric now. Split it at the top: 1.02 s up, rising 5.10 m above the cliff top.

Hint 2
From the top, the ball falls from rest through $15 + 5.10 = 20.10$ m. Use $20.10 = 4.9 t^{2}$ .

Full answer
Up: 1.02 s. Down from the top: $t = 20.10 \div 4.9 \approx 2.03$ s. Total $\approx 3.05$ s. Range $\approx 17.3 \times 3.05 \approx 52.8$ m. Check: this is further than the 35.3 m on level ground, as it should be, because the extra 15 m of fall gives the ball about 1 s more to travel sideways.

The bigger idea

This topic has used one idea three times. Split a motion into two directions at right angles. Solve each direction with the tools for a straight line. Join the answers with the shared time. Resolving a velocity with sine and cosine is the last step that makes the idea work for any launch.

The same splitting works for forces, which are also arrows with a size and a direction. The next topic, Newton's laws, asks not only how things move but why, and resolving forces into components will be one of its most useful tools.

Key concepts in this lesson

A launch velocity u at an angle θ above the horizontal resolves into a horizontal part u cos θ and a vertical part u sin θ.
For a launch and landing at the same height, the time to the top is u sin θ ÷ g, and the total time is twice that.
The greatest height depends only on the vertical part, and the range is the horizontal part times the total time.
On level ground with no air resistance, angles that add to 90° give the same range, and 45° gives the greatest range.
When the landing is higher or lower than the launch, split the flight at the top instead of doubling the time.