Question

A stunt motorcyclist is attempting to jump over a river of width 60m using a ramp which is inclined at 30 degrees to the horizontal. Assume the height of the ramp is small enough to be ignored in the calculations.
a) What is the minimum speed the stunt motorcyclist would need to jump over the river?
b) What it the maximum height above the river that the stunt motorcyclist reaches while jumping?

Answer 1

a. Range = Vo^2*sin(2A)/g = 60 m.
Vo^2*sin(60)/9.8 = 60
0.0884Vo^2 = 60
Vo^2 = 679
Vo = 26.06 m/s.[30o]

b. Yo = 26.06*sin30 = 13.03 m/s = Ver.
component.

hmax = (Y^2-Yo^2)/2g
hmax = (0-13.03^2)/-19.6 = 8.66 m.