Question

A motorcycle rider plans to make a jump off a ramp. If he leaves the ramp with a speed of 38.0m/s and has a speed of 36.3m/s at the top of his trajectory, determine his max height above the end of the ramp. Ignore friction and air resistance. i got 6.45m but i really don't know if I'm doing the question right...

Answer 1

u = constant horizontal velocity = 36.3
so
36.3 = 38 cos theta
so
theta = 17.2 degrees above horizontal

Vi = initial vertical velocity
= 38 sin theta = 38 sin 17.2 = 11.2 m/s

v = Vi - 9.81 t
at top v = 0
so
t = 11.2/9.81 = 1.15 seconds to top

h = Vi t - 4.9 t^2
= 11.2(1.15) - 4.9 (1.15)^2
= 12.8 - 6.4
= 6.4 meters
I think you did it right

Answer 2

Thank you!