Posted by jay on .
A quarterback throws the football to a stationary receiver who is 22.7 m down the field.The football is thrown at an initial angle of 39◦ to the ground.
The acceleration of gravity is 9.81 m/s2 .
a) At what initial speed must the quarter-
back throw the ball for it to reach the receiver?
S is the speed
horizontal speed = u = S cos 39
22.7 = u t
so t = 22.7/u
in the air to that time t
Vo = S sin 39
h = 0 + Vo t - 4.9 t^2
h = 0 at the end as well as at t = 0
Vo t = 4.9 t^2
t = 0 ( not a help) or t = Vo/4.9
Vo/4.9 = 22.7/u
22.7 *4.9 = 111 = Vo u
111 = S sin 39 * S cos 39
S^2 = 111/(sin 39 cos 39)
What is the ball’s highest point during its flight? please