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 -16t^{2 }-4 = (Vo sin45)( 114√2Vo) -16( 114√2Vo)^{2 }-4 = 114Vo^{2} -16( 25992)Vo^{2 }-4 = 114Vo^{2} -415872Vo^{2}-4 = (114-415872)Vo^{2 }-4 = -415758Vo^{2 }-4/-415758 = Vo^{2 }^{Vo= }√(2/207879)Vo = .003102
Possible answers;
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