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 ofwith the horizontal. Find the speed of the football when it is released. Round your answer to three decimal places.(Vo Cosθ)t

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

Vo cosθ t = x

Vo cos(45) t = 114

t=114/Vo cos45

t= 114√2Vo

[ h+(Vosinθ)t - ˝gt˛] = y4= 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