Projectile on an inclined plane
Thrown at 20 m/s and 60° up a 30° slope with g = 10 m/s², the ball is in the air for 2.31 s and lands 26.67 m up the slope.
Projectile motion
Drag the launch arrow, or type the speed and angle. Slide the time to see the velocity at that moment. The dots are equal steps of time apart. Air resistance is ignored, as in NCERT.
Time of flight T2.31 s2u sin(θ - α) / g cos α
Range on slope R26.67 malong the slope
Height above slope5.77 mu² sin²(θ - α) / 2g cos α
Velocity at 0.58 s15.28 m/svx = 10, vy = 11.55, 49.11° up
Pathy = 1.73x - 0.05x²y = x tan θ - gx² / (2u² cos²θ)
A ball thrown at 20 m/s and 60 degrees up a 30 degree slope lands 26.67 m up the slope after 2.31 s.
Show the formulas
- Time of flight
- T = 2u sin(θ - α) / (g cos α) = 2 × 20 × sin 30° / (10 × cos 30°) = 2.31 s
- Range on the slope
- R = 2u² sin(θ - α) cos θ / (g cos²α) = 2 × 20² × sin 30° × cos 60° / (10 × cos²30°) = 26.67 m
- Height above slope
- H = u² sin²(θ - α) / (2g cos α) = 5.77 m
- Longest range
- θ = 45° + α/2 = 60°, R max = u² / (g(1 + sin α)) = 26.67 m
- Velocity
- vx = u cos θ = 10 m/s, vy = u sin θ - gt = 11.55 m/s at t = 0.58 s, so v = 15.28 m/s
- Path
- y = x tan θ - gx² / (2u² cos²θ), here y = 1.73x - 0.05x²
Step by step
- Given u = 20 m/s at θ = 60° up a slope of α = 30°, with g = 10 m/s².
- The ball leaves at θ - α = 30° to the slope. Across the slope gravity is g cos α = 8.66 m/s².
- T = 2u sin(θ - α) / (g cos α) = 2 × 20 × sin 30° / (10 × cos 30°) = 2.31 s
- R = 2u² sin(θ - α) cos θ / (g cos²α) = 2 × 20² × sin 30° × cos 60° / (10 × cos²30°) = 26.67 m along the slope
- The greatest height above the slope is u² sin²(θ - α) / (2g cos α) = 5.77 m.
Why it works
- Take axes along the slope and at right angles to it. The launch makes θ - α with the slope.
- Gravity has a part g sin α down the slope and g cos α into it.
- Across the slope: the ball leaves at u sin(θ - α) and returns to the slope after T = 2u sin(θ - α) / (g cos α).
- Horizontally it moves at u cos θ, so it lands u cos θ × T away along the ground. Divide by cos α to measure along the slope: R = 2u² sin(θ - α) cos θ / (g cos²α).
- R is largest when 2θ - α = 90°, that is θ = 45° + α/2, and then R max = u² / (g(1 + sin α)).
Projectile on an inclined plane: quick answers
What is the time of flight on an inclined plane?
T = 2u sin(θ - α) / (g cos α), with θ the launch angle above the horizontal and α the slope angle.
What is the range up an inclined plane?
R = 2u² sin(θ - α) cos θ / (g cos²α), measured along the slope. Here it is 26.67 m.
Why is the angle θ - α used?
Across the slope, the ball leaves at u sin(θ - α) and gravity pulls it back at g cos α. That decides when it touches the slope again.