posted by Abdullah .
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.
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