Physics in 2 dimensions-dynamics?

Question

A motorcycle daredevil wants to set a record for jumping over burning school buses. He has hired you to help with the design. He intends to ride off a horizontal platform at 40 m/s, cross the burning buses in a pit below him, then land on a ramp sloping down at 20 degrees. It's very important that he not bounce when he hits the landing ramp because that could cause him to lose control and crash. You immediately recognize that he won't bounce if his velocity is parallel to the ramp as he touches down. This can be accomplished if the ramp is tangent to his trajectory and if he lands right on the front edge of the ramp. There's no room for error! Your task is to determine where to place the landing ramp. That is, how far from the edge of the launching platform should the front edge of the landing ramp be horizontally and how far below it?

How far from the edge of the launching platform should the front edge of the ramp be horizontally?
How far below the launching platform should the front edge of the ramp be?

Answer 1

I will use the flight time to base the calculations.

Note that air resistance is being ignored - poor devil.

Any way.

The distance the rider will fly is d
d=40*t
where t is the flight time

In flight, gravity will act on the rider and bike so that at time t, his downward velocity will be:

g*t, where g=9.81

In order for his trajectory to be 20 degrees from horizontal requires that
Tan(20)=g*t/40
so
t=40*tan(20)/g
=1.484 sec

now that I know t, I can compute d
as 40*1.484
=59.4 m

The drop will be
1/2 * g*t^2
=10.8 m

j