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.
Answers
Steve
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
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
B
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