A body is projected with a velocity of `40 ms^(-1)`. After `2 s` it crosses a vertical pole of height ` 20 .4 m` Find the angle of projection and horizontal range of projectile. (g = 9.8 ms^(-2)`.

Options

A. ` 15^@`
B. ` 30^@`
C. ` 45^@`
D. ` 60^@`

Answer

B

Explanation

Here, ` u = 40 m//s , BA = 20 .4 m. Let ` theta` the angle of projection at (O). Taking vertical motion or projectile from (O) to (A) we have
` .
` u = u sin theta = 40 sin theta , a =- 9.8 m//s^@`,
` S= 20 . 4 m, t = 2 s. As ` S= ut + 1/2 a t^2`
:. ` 20 .4 = ( 40 sin theta) xx 2 + 1/2 (- 9.8 a t^2`
or ` 20 .4 = 80 sin theta - 19 .6`
or ` 80 sin theta = 20 .4 + 19 .6 = 40`
or ` sin theta = (40)/(80) = 1/2 = sin 30^@` or ` hteta = 30^@`.