Question

A daredevil on a motorcycle leaves the end of a ramp with a speed of 34.5 m/s as in the figure below. If his speed is 32.1 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance.

Answer 1

well, if his horzontal velocity is 32.1m/s, and his total at launch is 34.5m/s,Then his initial vertical velocity is sqrt (34.5^2-32.1^2)=12.6m/s check that.

Vi^2=2gh
solve for h.

Answer 2

The 32.11 m/s is the unchanging horizontal component. That and Mr. Pythagoras will tell you the initial vertical component.

sqrt[(35.5)^2 - (32.1)^2] = Vyo

From Vyo, you can derive the maximum height.

Go for it!