Question

I need help with this physics on projectile motion problem!

4.83: 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 40m/s, cross the burning buses in a pit below him, then land on a ramp sloping down at 20∘. 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?

And!

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

Thanks for the help! You guys rock!

Answer 1

angle made by the final velocity with the horizantal =20

tan20 = vy/vx)

tan20 = (vy/40)

vy = 14.6 m/s

final vertical component = vy = 14.6 m/s

along vertical

vy = voy + ay*t

14.6 = 0 9.8*t

t = 1.5 s

horizantal distance = x = vx*t = 40*1.5 = 60 m

vertical distance = y = 0.5*g*t^2 = 0.5*9.8*1.5^2 = 11.025 m