B. Kinematics · Kinematics: Motion in One and Two Dimensions

Projectiles Launched at an Angle: Airtime, Maximum Height and Range

B2.8 B2.9 B3.3 B2.7About 75 minutes

Learning goals

I can resolve a launch velocity into horizontal and vertical components.
I can calculate the time of flight, maximum height and range of a projectile launched at an angle.
I can explain how the launch angle affects airtime, maximum height and range, and why 45° gives the greatest range on level ground.
I can solve projectile problems where the launch and landing heights differ.

Words to know

launch angle (θ)
The angle between the initial velocity and the horizontal.

maximum height
The highest point of the trajectory, where the vertical velocity is zero.

symmetry of a trajectory
On level ground (no air resistance), the time up equals the time down, and the landing speed equals the launch speed.

complementary angles
Two angles that add to 90°, such as 30° and 60°. On level ground they give the same range.

The perfect kick

A soccer player taking a long kick, a golfer hitting a drive, a basketball player shooting a free throw: each chooses a launch angle and a speed. Too flat and the ball hits the ground early; too steep and it goes high but not far. In this lesson you will predict exactly how high, how long and how far, and find the best angle.

Start by resolving the launch velocity

When an object is launched at speed v₁ and angle θ above the horizontal, split v₁ into components:

v₁x = v₁ cos θ (horizontal; stays constant)
v₁y = v₁ sin θ (vertical; changes at 9.8 m/s² downward)

Then treat the two directions separately, just like last lesson. Take up as positive for the vertical motion, so ay = −9.8 m/s².

A soccer ball kicked at 18 m/s at 30° above the horizontal. Its horizontal component is 15.6 m/s and its vertical component is 9.0 m/s.

Three key results (level ground)

Maximum height: at the top, vy = 0. Use v₂y² = v₁y² + 2ayΔdy: H = v₁y² ÷ (2 × 9.8). It depends only on the vertical component.
Time of flight: time to the top is v₁y ÷ 9.8, and the time down is the same, so Δt = 2v₁y ÷ 9.8. Again, only the vertical component matters.
Range: horizontal motion is uniform, so Δdx = v₁xΔt.

At the top of the path the ball is still moving: its velocity is v₁x, horizontal. Its acceleration is still 9.8 m/s² [down].

Worked example

Example 1: A soccer kick

A soccer ball is kicked from level ground at 18 m/s at 30° above the horizontal. Find (a) the time of flight, (b) the maximum height, and (c) the range.

Components: v₁x = 18 cos 30° = 15.59 m/s; v₁y = 18 sin 30° = 9.00 m/s.
(a) Time to the top: v₂y = v₁y + ayΔt → 0 = 9.00 − 9.8Δt → Δt = 0.918 s. Total: Δt = 1.84 s. (Or use Δdy = 0: 0 = 9.00Δt − 4.9Δt², so Δt = 9.00 ÷ 4.9 = 1.84 s.)
(b) H = v₁y² ÷ 19.6 = 81.0 ÷ 19.6 ≈ 4.1 m.
(c) Δdx = v₁xΔt = 15.59 × 1.837 ≈ 28.6 m.

(a) 1.84 s; (b) 4.1 m; (c) 28.6 m

Worked example

Example 2: A golf drive at 45°

A golf ball leaves the tee at 30 m/s at 45° above the horizontal. Ignoring air resistance, find the airtime, maximum height and range. What is its velocity at the top, and when it lands?

v₁x = 30 cos 45° = 21.21 m/s; v₁y = 30 sin 45° = 21.21 m/s.
Airtime: Δt = 2 × 21.21 ÷ 9.8 ≈ 4.33 s.
Maximum height: H = 21.21² ÷ 19.6 ≈ 23.0 m.
Range: Δdx = 21.21 × 4.329 ≈ 91.8 m.
At the top: vy = 0, so v = 21.2 m/s horizontally. On landing: by symmetry vy = 21.2 m/s [down], so v = 30 m/s at 45° below the horizontal.

4.33 s; 23.0 m; 91.8 m; 21.2 m/s horizontal at the top; 30 m/s at 45° below the horizontal on landing

The golf drive of Example 2 (to scale). The horizontal arrow is the same everywhere. The vertical arrow shrinks to zero at the top and grows again on the way down, so the path is symmetrical.

Which angle goes farthest?

Put the three results together for level ground:

Δdx = v₁xΔt = (v₁ cos θ)(2v₁ sin θ ÷ g) = v₁²(2 sin θ cos θ) ÷ g

A trigonometric identity (you will meet it in MHF4U) says 2 sin θ cos θ = sin 2θ, so Δdx = v₁² sin 2θ ÷ g. Since sin 2θ is largest (equal to 1) when 2θ = 90°:

The maximum range on level ground, without air resistance, happens at θ = 45°.
Complementary angles (such as 30° and 60°, or 15° and 75°) give the same range, because sin 60° = sin 120° and so on. The steeper launch goes higher and stays up longer.

You can check this with Example 1: kicking at 60° instead of 30° gives v₁x = 9.00 m/s and v₁y = 15.59 m/s, an airtime of 3.18 s and a range of 9.00 × 3.18 = 28.6 m, the same as before, but a height of 12.4 m.

In real life, air resistance shortens every trajectory, and the best angle is usually a little less than 45°. Launching from a height also lowers the best angle.

Trajectories for a launch speed of 20 m/s at five angles (to scale, no air resistance). 45° goes farthest. 30° and 60° land at the same spot, and so do 15° and 75°.

Worked example

Example 3: Launched from a cliff

A ball is thrown at 15 m/s at 30° above the horizontal from the top of a cliff 20 m above the water. Find (a) its time of flight and (b) its horizontal distance from the cliff when it lands.

v₁x = 15 cos 30° = 12.99 m/s; v₁y = 15 sin 30° = 7.50 m/s. Up positive; the water is at Δdy = −20 m.
Vertical: Δdy = v₁yΔt + ½ayΔt² → −20 = 7.50Δt − 4.9Δt².
Standard form: 4.9Δt² − 7.50Δt − 20 = 0. Quadratic formula: Δt = [7.50 ± √(7.50² + 4 × 4.9 × 20)] ÷ (2 × 4.9) = [7.50 ± √448.25] ÷ 9.8 = (7.50 ± 21.17) ÷ 9.8.
The positive root is Δt ≈ 2.93 s (the negative root is before the throw, so reject it).
(b) Δdx = 12.99 × 2.926 ≈ 38 m.
Note: the symmetry shortcut (Δt = 2v₁y ÷ g) works only when launch and landing heights are equal. Here you must use the full equation.

(a) 2.93 s; (b) about 38 m

Just for fun

Golf balls have dimples for a reason: the dimples change how air flows around the spinning ball and help it travel much farther than a smooth ball would. Real golf is one sport where "ignore air resistance" would give you a very wrong answer!

Hands-on activity

Investigate launch angle with a water stream

You need: a garden hose with an adjustable nozzle on a warm day (or a squeeze bottle or spray bottle over a lawn or driveway), a measuring tape, a protractor (or a phone angle app), and a partner. Grown-up note: do this outdoors where water won't cause slipping or damage, and never aim at people, windows or power lines.

Hold the nozzle at a fixed height (say, on the ground or at a marked height) and keep the water pressure the same for every trial.
Your partner measures the launch angle with the protractor held beside the nozzle. Try 15°, 30°, 45°, 60° and 75°.
For each angle, measure the range (where the stream hits the ground) and estimate the maximum height of the stream against a wall or pole.
Graph range against angle. Which angle went farthest? Do 30° and 60° (or 15° and 75°) give similar ranges?
How do height and airtime change with angle?

What to look for: the range should be greatest near 45° (often a little lower, because air resistance breaks up the stream), and complementary angles should give similar ranges. Maximum height should increase steadily with angle. If your nozzle is above the ground, the best angle shifts slightly below 45°.

Your turn

A ball is launched at 25 m/s at 40° above the horizontal. Find v₁x and v₁y.

For that ball (level ground), find the time of flight.

Find its maximum height and range.

What is the ball's speed at the top of its path?

A ball launched at 20° has a range of 31.7 m. What other launch angle (same speed, level ground) gives the same range?

Which launch angle gives the longest airtime for a fixed launch speed? Explain.

More practice: warm-up

At the top of its trajectory, what is a projectile's vertical velocity?

Which component of the launch velocity determines the maximum height?

On level ground with no air resistance, which angle gives the maximum range?

A ball is launched with v₁y = 9.8 m/s. How long does it take to reach the top?

Which angle pairs with 25° to give the same range?

More practice: on your own

A basketball is shot at 12 m/s at 50° and caught at the same height. Find its airtime and horizontal distance.

A football is kicked at 16 m/s at 35°. Find its maximum height.

A ball is kicked with v₁x = 12 m/s and v₁y = 5.0 m/s. Find its launch speed and angle.

Use Δdx = v₁² sin 2θ ÷ g to find the range of a ball launched at 20 m/s at 30°. Check with the component method.

Two balls are launched at the same speed, one at 30° and one at 60°. Compare their ranges, airtimes and maximum heights.

Challenge

In Example 3 (15 m/s at 30° from a 20 m cliff), find the ball's speed just before it hits the water.

A ball launched at 45° on level ground has a range of 40 m. Find its launch speed.

A long jumper leaves the ground at 9.0 m/s at 20°. Treating her as a projectile landing at the same height, how far does she jump? Why do real long jumpers use angles around 20° rather than 45°?

Check your understanding

Quiz

A ball is launched at 20 m/s at 30°. Its horizontal component is:

At the highest point of a projectile's path, its velocity is:

On level ground (no air resistance), the airtime of a projectile depends on:

A projectile launched at 70° has the same range as one launched (same speed) at:

A ball with v₁y = 14.7 m/s is launched from level ground. Its airtime is:

Why does a real golf drive or water stream usually go farthest at an angle a little below 45°?

Which statement about a projectile launched at an angle on level ground is true?

For parents and teachers

Big idea: For a projectile launched at an angle, the vertical component alone sets the airtime and maximum height, and the horizontal component times the airtime sets the range (B2.8, B2.9). On level ground, 45° gives the greatest range and complementary angles give equal ranges (B3.3). The golf ball at 45° is an example named in the curriculum.

How to help: The water-stream activity makes the angle pattern visible. Ask, "Which part of the velocity decides how long it stays up? Which part decides how far it goes each second?"

Common difficulties:

Using the full launch speed in vertical or horizontal equations instead of its components.
Using Δt = 2v₁y ÷ g when launch and landing heights differ (use the quadratic instead).
Saying the velocity is zero at the top (only the vertical part is).
Calculator in radian mode.