A Motorcycle Jump. You are planning to make a jump with your motorcycle by driving over a ramp that will launch you at an angle of 30.0o with respect to the horizontal. The front edge of the ramp on which you are supposed to land, however, is 25.0 ft lower than the edge of the launch ramp (i.e., your launch height).
(a) Assuming a launch speed of 70.0 mph, at what horizontal distance from your launch point should the landing ramp be placed?
(b) In order to land smoothly, the angle of the landing ramp should match the direction of your velocity vector when you touch down. What should be the angle of the landing ramp?
70 mi/hr = 102.67 ft/s
your horizontal speed is 102.67 * √3/2 = 88.9 ft/s
your vertical speed at time t seconds is 102.67 * 1/2 - 32t
so, launching from a height of 25 ft, it will take t seconds to get
25 + 51.335t - 16t^2 = 0
t = 3.638 seconds to land on the ramp.
your horizontal distance is thus 3.638 * 88.9 = 323.418 ft
your vertical speed is -65.081 ft/s
Thus the angle θ is such that
tanθ = 65.081/88.9 = 36.2°