# Thread: Find the speed of the football when it is released.

1. ## Find the speed of the football when it is released.

What am I doing wrong...

The quarterback of a football team releases a pass at a height of 8 feet above the playing field, and the football is caught by a receiver 38 yards directly downfield at a height of 4 feet. The pass is released at an angle of with the horizontal. Find the speed of the football when it is released. Round your answer to three decimal places.

Formula used:
(Vo Cosθ)t i + [ h+(Vosinθ)t - ˝gt˛] j

OK so. My method to solve this was to solve the X portion for t with respect to Vo and then plug t into the y portion to solve for Vo. 38 yards = 114 ft

Vo cosθ t = x
Vo cos(45) t = 114
t=114/Vo cos45
t= 114√2Vo

[ h+(Vosinθ)t - ˝gt˛] = y
4= 8+(Vo sin45)t -16t2
-4 =
(Vo sin45)( 114√2Vo) -16( 114√2Vo)2
-4 = 114Vo2 -16( 25992)Vo2
-4 = 114Vo2 -415872Vo2-4 = (114-415872)Vo2
-4 = -415758Vo2
-4/-415758 = Vo2
Vo= √(2/207879)Vo = .003102

 a. 19.462 feet per second b. 33.169 feet per second c. 118.732 feet per second d. 59.366 feet per second e. 190.165 feet per second

What am I doing wrong!!
This is killing me and i have quiz on it tomorrow

2. ## Re: Find the speed of the football when it is released.

Originally Posted by shmiks

The quarterback of a football team releases a pass at a height of 8 feet above the playing field, and the football is caught by a receiver 38 yards directly downfield at a height of 4 feet. The pass is released at an angle of with the horizontal. Find the speed of the football when it is released. Round your answer to three decimal places.
Check the units... Otherwise the solution method is sound.

-Dan