The equations of motion of a projectile are given by x = 36t metre and `2y = 96t - 9.8r^(2)` metre. The angle of projection is :

Options

A. `sin^(-1)(4/5)`
B. `sin^(-1)(3/5)`
C. `sin^(4/3)`
D. `sin^(-1)(3/4)`

Answer

A

Explanation

Here `v_(x) -dx/dt=36, v_(y)=(dy)/(dt)=48-9.8t`
for `t=0v_(y)=48`
`:.v=(36^(2)+(48)^(2))^(1//2)=60`
Now `sintheta=v_(x)/v=48/60=4/5`
`theta=sin^(-1)(4/5)`