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 = 36 t and 2y = 96 t - 9.8 t^2`
or `y = 48t - 4.9 t^(2)`
Let the initial velocity of projectile be u and angle of projection be `theta` . Then
initial horizontal component of velocity,
`u_x=ucostheta=((dx)/(dt))=36`
i.e., `u cos theta = 36 " "..(i)`
initial vertical component of velocity ,
`u_y = usin theta =((dy)/(dt)) = 48 ` i.e, `usin theta = 48 " "...(ii)`
From (i) and (ii) ,
`tantheta=48/36 =4/3 , :.sin theta=4/5" or " theta = sin^(-1) (4/5)`