A BMX rider is coming up to a ramp that is set to 23.0° and the end point is 1.60 m above the ground. They travel up the ramp at a steady speed of 9.20 m/s. What speed will they have just before they hit the ground?

Answer 1

Starting at the top of the ramp, the rider's height is

h(t) = 1.60 sin23° t - 4.9t^2

So, find t when h=0

The rider's velocity when he hits the ground is

Vx = 1.60 cos23°
Vy = 1.60 sin23° - 9.8t
So, the speed v at time t is found using
v^2 = Vx^2 + Vy^2