Physics · Simulation and virtual lab Free
Projectile motion: launch speed, angle and range
Launch a ball at different angles and speeds, watch its path every 0.1 s, and find which angle throws it furthest.
- You will learn to
- Measure the time of flight, range and maximum height of a projectile.
- Explain why horizontal velocity is constant while vertical velocity changes.
- Find the launch angle for maximum range, from the ground and from a height.
- You will measure
- Time of flight (s), range (m) and highest point (m) for each angle, speed and height.
- Time needed
- About 30–40 min
- Equipment
- Nothing extra. Real version: marble and ramp at the edge of a bench, metre rule, carbon paper, or a phone video of a thrown ball.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To investigate how the launch angle, launch speed and launch height affect the time of flight, range and maximum height of a projectile, and to find the angle that gives the maximum range.
What to do
- Keep speed 12 m/s, height 0 m, Kigali. Set the angle to 15° and tap Launch the ball, then Record this launch.
- Repeat for 30°, 45°, 60° and 75°. Look at the graph of range against angle.
- Which angle gives the longest range? Which two angles give the same range?
- Use the slow-motion film: the green arrow (vx) stays the same length but the red arrow (vy) shrinks, becomes zero at the top and grows downwards. Explain why.
- Set the height to 5 m and repeat 30°, 35°, 40°, 45°. Is 45° still the best angle?
- Choose the Moon and launch at 45°. Compare the range with Kigali and explain the difference.
Results
Questions
What is a projectile?
An object thrown into the air that then moves only under the action of gravity (air resistance ignored).
Why does the horizontal velocity stay constant?
No horizontal force acts on the ball (no air resistance), so by Newton’s first law its horizontal velocity does not change.
What is the vertical velocity at the highest point?
Zero: the ball stops rising and starts falling. Only the horizontal velocity remains.
A ball is launched at 12 m/s at 45° on level ground (g = 9.78 m/s²). Calculate its range.
R = u² sin 2θ ÷ g = 144 × 1 ÷ 9.78 = 14.7 m.
Why do 30° and 60° give the same range on level ground?
Range depends on sin 2θ, and sin 60° = sin 120°; complementary angles give the same value. The 60° ball goes higher but more slowly forward, for longer.
Why is the range much longer on the Moon?
g on the Moon (1.62 m/s²) is about 6 times smaller, so the ball stays in the air about 6 times longer and travels about 6 times further.
For teachers: expected results and how the model works
Exact equations (no air resistance): T = (u sin θ + √(u² sin² θ + 2gh)) ÷ g, R = u cos θ × T, H = h + u² sin² θ ÷ 2g. g = 9.78 m/s² in Kigali, 1.62 m/s² on the Moon. Check values at 12 m/s, h = 0: 45° gives T = 1.74 s, R = 14.72 m, H = 3.68 m; 30° and 60° both give R = 12.75 m (H = 1.84 m and 5.52 m). At 20 m/s, 45°: R = 40.9 m. From h = 5 m at 12 m/s: 35° gives 19.03 m, 40° gives 19.04 m, 45° gives 18.67 m, so the best angle is below 45°. Moon at 12 m/s, 45°: R = 88.9 m. Dots are drawn every 0.1 s: equal horizontal spacing shows constant vx.
Exam practice Exam set
National-exam style questions on this simulation (20 marks). Try each one on paper first, then open the marking scheme.
A stone is thrown horizontally at 8 m/s from the top of a cliff 20 m high (g = 9.8 m/s²). Calculate (a) the time to reach the ground, (b) the horizontal distance travelled, (c) its speed on landing. (7 marks)
Marking scheme: (a) 20 = ½ × 9.8 × t² (1), t = 2.02 s (1). (b) x = 8 × 2.02 = 16.2 m (2). (c) vy = 9.8 × 2.02 = 19.8 m/s (1); v = √(8² + 19.8²) (1) = 21.4 m/s (1).
A ball is kicked at 20 m/s at 30° above the horizontal from level ground (g = 9.8 m/s²). Find (a) the time of flight, (b) the maximum height, (c) the range. (6 marks)
Marking scheme: uy = 20 sin 30° = 10 m/s and ux = 20 cos 30° = 17.3 m/s (1). (a) T = 2 × 10 ÷ 9.8 = 2.04 s (2). (b) H = 10² ÷ (2 × 9.8) = 5.1 m (1). (c) R = 17.3 × 2.04 = 35.3 m (2).
Show that the range on level ground is R = u² sin 2θ ÷ g and state the angle for maximum range. (4 marks)
Marking scheme: T = 2u sin θ ÷ g (1); R = u cos θ × T = 2u² sin θ cos θ ÷ g (1) = u² sin 2θ ÷ g (1); maximum when sin 2θ = 1, θ = 45° (1).
State two assumptions made in these calculations and say how real air would change the range. (3 marks)
Marking scheme: No air resistance (1); g constant / the Earth is flat over the range (1); air resistance reduces the range (and the best angle becomes less than 45°) (1).
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Projectile motion: launch speed, angle and range
Aim: To investigate how the launch angle, launch speed and launch height affect the time of flight, range and maximum height of a projectile, and to find the angle that gives the maximum range.
Apparatus (real laboratory): marble and ramp at the edge of a bench, metre rule, carbon paper, or a phone video of a thrown ball.
What to do (with the simulation on www.donatstem.co.rw)
- Keep speed 12 m/s, height 0 m, Kigali. Set the angle to 15° and tap Launch the ball, then Record this launch.
- Repeat for 30°, 45°, 60° and 75°. Look at the graph of range against angle.
- Which angle gives the longest range? Which two angles give the same range?
- Use the slow-motion film: the green arrow (vx) stays the same length but the red arrow (vy) shrinks, becomes zero at the top and grows downwards. Explain why.
- Set the height to 5 m and repeat 30°, 35°, 40°, 45°. Is 45° still the best angle?
- Choose the Moon and launch at 45°. Compare the range with Kigali and explain the difference.
Results
| # | Angle (°) | Speed (m/s) | Height (m) | Time of flight (s) | Range (m) | Highest point (m) |
|---|
| 1 | | | | | | |
| 2 | | | | | | |
| 3 | | | | | | |
| 4 | | | | | | |
| 5 | | | | | | |
| 6 | | | | | | |
| 7 | | | | | | |
| 8 | | | | | | |
Graph
Plot range (m) on the vertical axis against launch angle (°). Mark the angle of maximum range.
Questions
- What is a projectile? [2]
- Why does the horizontal velocity stay constant? [2]
- What is the vertical velocity at the highest point? [2]
- A ball is launched at 12 m/s at 45° on level ground (g = 9.78 m/s²). Calculate its range. [2]
- Why do 30° and 60° give the same range on level ground? [2]
- Why is the range much longer on the Moon? [2]
- A stone is thrown horizontally at 8 m/s from the top of a cliff 20 m high (g = 9.8 m/s²). Calculate (a) the time to reach the ground, (b) the horizontal distance travelled, (c) its speed on landing. [7]
- A ball is kicked at 20 m/s at 30° above the horizontal from level ground (g = 9.8 m/s²). Find (a) the time of flight, (b) the maximum height, (c) the range. [6]
- Show that the range on level ground is R = u² sin 2θ ÷ g and state the angle for maximum range. [4]
- State two assumptions made in these calculations and say how real air would change the range. [3]
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Projectile motion: launch speed, angle and range
Expected results
Exact equations (no air resistance): T = (u sin θ + √(u² sin² θ + 2gh)) ÷ g, R = u cos θ × T, H = h + u² sin² θ ÷ 2g. g = 9.78 m/s² in Kigali, 1.62 m/s² on the Moon. Check values at 12 m/s, h = 0: 45° gives T = 1.74 s, R = 14.72 m, H = 3.68 m; 30° and 60° both give R = 12.75 m (H = 1.84 m and 5.52 m). At 20 m/s, 45°: R = 40.9 m. From h = 5 m at 12 m/s: 35° gives 19.03 m, 40° gives 19.04 m, 45° gives 18.67 m, so the best angle is below 45°. Moon at 12 m/s, 45°: R = 88.9 m. Dots are drawn every 0.1 s: equal horizontal spacing shows constant vx.
Answers
- What is a projectile? [2]
An object thrown into the air that then moves only under the action of gravity (air resistance ignored).
- Why does the horizontal velocity stay constant? [2]
No horizontal force acts on the ball (no air resistance), so by Newton’s first law its horizontal velocity does not change.
- What is the vertical velocity at the highest point? [2]
Zero: the ball stops rising and starts falling. Only the horizontal velocity remains.
- A ball is launched at 12 m/s at 45° on level ground (g = 9.78 m/s²). Calculate its range. [2]
R = u² sin 2θ ÷ g = 144 × 1 ÷ 9.78 = 14.7 m.
- Why do 30° and 60° give the same range on level ground? [2]
Range depends on sin 2θ, and sin 60° = sin 120°; complementary angles give the same value. The 60° ball goes higher but more slowly forward, for longer.
- Why is the range much longer on the Moon? [2]
g on the Moon (1.62 m/s²) is about 6 times smaller, so the ball stays in the air about 6 times longer and travels about 6 times further.
- A stone is thrown horizontally at 8 m/s from the top of a cliff 20 m high (g = 9.8 m/s²). Calculate (a) the time to reach the ground, (b) the horizontal distance travelled, (c) its speed on landing. [7]
(a) 20 = ½ × 9.8 × t² (1), t = 2.02 s (1). (b) x = 8 × 2.02 = 16.2 m (2). (c) vy = 9.8 × 2.02 = 19.8 m/s (1); v = √(8² + 19.8²) (1) = 21.4 m/s (1).
- A ball is kicked at 20 m/s at 30° above the horizontal from level ground (g = 9.8 m/s²). Find (a) the time of flight, (b) the maximum height, (c) the range. [6]
uy = 20 sin 30° = 10 m/s and ux = 20 cos 30° = 17.3 m/s (1). (a) T = 2 × 10 ÷ 9.8 = 2.04 s (2). (b) H = 10² ÷ (2 × 9.8) = 5.1 m (1). (c) R = 17.3 × 2.04 = 35.3 m (2).
- Show that the range on level ground is R = u² sin 2θ ÷ g and state the angle for maximum range. [4]
T = 2u sin θ ÷ g (1); R = u cos θ × T = 2u² sin θ cos θ ÷ g (1) = u² sin 2θ ÷ g (1); maximum when sin 2θ = 1, θ = 45° (1).
- State two assumptions made in these calculations and say how real air would change the range. [3]
No air resistance (1); g constant / the Earth is flat over the range (1); air resistance reduces the range (and the best angle becomes less than 45°) (1).