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)(4/3)`
D. `sin^(-1)(3/4)`

Answer

a

Explanation

Given x=36t
and `2y=96t-9.8t^(2)`
or `y=48t-4.9t^(2)`
Let the initial velocity of projectile be u and angle of projection `theta`. Then, Initial horizontal component of velocity,
`u_(x)=ucos theta=((dx)/(dt))_(t=0)=48`...........(i)
or `u sin theta=48`.......(ii)
Dividing (ii) by (i), we get
`:. tan theta=48/36=4/3`
`sin theta=4/5 or theta=sin^(-1)(4/5)`