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 https://fsulearn.ferris.edu/courses/...es/mc021-1.jpgwith 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
[ 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
Re: Find the speed of the football when it is released.
Check the units... Otherwise the solution method is sound.
Originally Posted by shmiks