Question

The quarterback of a football team releases a pass at a height of 7 feet above the playing field, and the football is caught at a height of 4 feet, 30 yards directly downfield. The pass is released at an angle of 35 degrees with the horizontal. The parametric equations for the path of the football are given by x=0.82vot and y=7+0.57vot-16t^2 where vo is the speed of the football (in feet per second) when it is released. Find the speed of the football when it is released.

Answer 1

I'll just use v for readability. We need to solve

Since the ball was caught 30 yards (90 feet) downfield,

.82vt = 90
t = 109.76/v

Now, knowing what t is in terms of v, we can solve

7+0.57v(109.76/v)-16(109.76/v)^2 = 4
v = 54.22

Answer 2

.4067