The horizontal distance, in feet, of a projectile that is fired with an initial velocity `upsilon`, in feet pet second, at an angle `theta` with the horizontal, is given by
`H(upsilon, theta)=(upsilon^(2)sin(2theta))/(32)`
If a football is kicked at an angle of 50 degrees with the horizontal and an initial velocity of 30 feet per second, what is the horizontal distance, in feet, from the point where the football is kicked to the point where the football first hits the ground ?

Options

A. 28
B. 30
C. 33
D. 36

Answer

A

Explanation

Just plug `theta = 50` and `upsilon = 30` into the formula and crank out the answer :
`H = (upsilon^(2)sin(2theta))/(32)`
`=((30^(2))sin(2xx50^(@)))/(32)`
`=(900 sin 100^(@))/(32)`
`~~28`