A daredevil on a motorcycle leaves the end of a ramp with a speed of 33.0 m/s as in the figure below. If his speed is 31.3 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

Since the horizontal component of his speed stays the same (31.3 m/s in this case), the vertical component (which is zero at maximum height) must initially be sqrt(33.0^2 - 31.3^2) = 10.46 m/s. The time spent going up is
T = (10.46 m/s)/g = 1.07 s

The height that it reaches is
H = Vy,average* T
= (Vy,initial/2)*T = 5.8 meters