A man on a motorcycle plans to make a jump as shown in the figure. If he leaves the ramp with a speed of 29.5 m/s and has a speed of 28.0 m/s at the top of his trajectory, determine his maximum height (h) above the end of the ramp. Ignore friction and air resistance.

Answer 1

I do not see any figure.

His speed at the top is his horizontal speed, u, for the whole trip

u = 28 m/s

at takeoff v^2 = Vi^2 + u^2
where v is total speed
and Vi is initial speed up
so
29.5^2 = 28^2 + Vi^2
Vi = 9.29 m/s
that is all I need to do the vertical problem
easy with energy
m g h = .5 m Vi^2
h = .5 Vi^2/9.81
h = 4.4 meters