The vertical height of the projectile at the time is given by `y=4t-t^(2)` and the horizontal distance covered is given by `x=3t`. What is the angle of projection with the horizontal ?

Options

A. `37^(@)`
B. `53^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

B

Explanation

(b): `(dx)/(dt)=3 and (dy)/(dt)=44t`
Substititing t=0, we can get the components of initial velocity along X-and Y-axis as follows:
`u_(x)=3m//s and u_(y)=4 m//s`
If `theta` is the angle of projection, then
`tan theta =(u_(y))/(u_(x))=(4)/(3)rArr theta=53^(@)`
Hence, option (b) is correct .