Numford
Projectile motion

Maximum range on an inclined plane

Up a 30° slope the best angle is 45° + α/2 = 60°. At 30 m/s with g = 10 m/s² the longest range is u² / (g(1 + sin α)) = 60 m.

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 T3.46 s2u sin(θ - α) / g cos α
Range on slope R60 malong the slope
Height above slope12.99 mu² sin²(θ - α) / 2g cos α
Velocity at 0.87 s22.91 m/svx = 15, vy = 17.32, 49.11° up
Pathy = 1.73x - 0.0222x²y = x tan θ - gx² / (2u² cos²θ)

A ball thrown at 30 m/s and 60 degrees up a 30 degree slope lands 60 m up the slope after 3.46 s.

g (m/s²)
Show the formulas
Time of flight
T = 2u sin(θ - α) / (g cos α) = 2 × 30 × sin 30° / (10 × cos 30°) = 3.46 s
Range on the slope
R = 2u² sin(θ - α) cos θ / (g cos²α) = 2 × 30² × sin 30° × cos 60° / (10 × cos²30°) = 60 m
Height above slope
H = u² sin²(θ - α) / (2g cos α) = 12.99 m
Longest range
θ = 45° + α/2 = 60°, R max = u² / (g(1 + sin α)) = 60 m
Velocity
vx = u cos θ = 15 m/s, vy = u sin θ - gt = 17.32 m/s at t = 0.87 s, so v = 22.91 m/s
Path
y = x tan θ - gx² / (2u² cos²θ), here y = 1.73x - 0.0222x²
Working

Step by step

  1. Given u = 30 m/s at θ = 60° up a slope of α = 30°, with g = 10 m/s².
  2. The range up the slope is R = 2u² sin(θ - α) cos θ / (g cos²α). It is largest when θ = 45° + α/2 = 45° + 15° = 60°.
  3. At θ = 60°: R = 2 × 30² × sin 30° × cos 60° / (10 × cos²30°) = 60 m.
  4. Check with R max = u² / (g(1 + sin α)) = 30² / (10 × (1 + sin 30°)) = 60 m.
  5. On level ground the same throw at 45° would go u² / g = 90 m, so the slope cuts the range.
Derivation

Why it works

  1. Take axes along the slope and at right angles to it. The launch makes θ - α with the slope.
  2. Gravity has a part g sin α down the slope and g cos α into it.
  3. 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).
  4. 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).
  5. R is largest when 2θ - α = 90°, that is θ = 45° + α/2, and then R max = u² / (g(1 + sin α)).
Questions

Range on an inclined plane: quick answers

At what angle is the range up an inclined plane maximum?

At θ = 45° + α/2 above the horizontal, which bisects the angle between the slope and the vertical.

What is the maximum range up a slope?

R max = u² / (g(1 + sin α)). At 30 m/s up a 30° slope it is 60 m.

What is the maximum range down a slope?

Throwing down the slope, the best angle is 45° - α/2 and R max = u² / (g(1 - sin α)), which is longer than on level ground.

Try your own throw