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

Options

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

Answer

A

Explanation

It is given that, `x = 36t`
`:. v_(x) = (dx)/(Dt) = 36 ms^(-1) rarr y = 48t - 4.9 t^(2)`
`:. v_(y) = 48 - 9.8 t`
at `t = 0 v_(x) = 36 ms^(-1)`
and `v_(y) = 48 ms^(-1)`
So, angle of projection, `theta = tan^(-1) ((v_(y))/(v_(x))) = tan^(-1) ((4)/(3))`
or `theta = sin^(-1) ((4)/(5))`