A ball must cross 43m to clear a tower and land. If the launch ramps inclined 20deg above the horizontal, what must the initial velocity be to cross safely?

Question

A ball must cross a horizontal distance of 43m to clear a tower and land safely. If the launch ramp is inclined 20 degrees above the horizontal, what must the initial velocity be to cross safely?Air resistance is negligible and the launch ramp and the landing ramp are the same height

Answer 1

You can use the range equation:

R = v₀²sin(2θ)/g ⇒

 

v₀ = √[Rg/sin(2θ)]

 

2θ = 2(20°) = 40°

R = 43 m

g = 9.8 m/s²

 

You want the range R to be at least 43 m, so the equation above establishes a minimum required speed.

 

v_min = √[(43)(9.8)/sin40°] m/s = 25.6 m/s