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

Explanation

Here, `u=40 ms^(-1),
`t=2 s, y =20 .4 m`
Taking vertical upward motion of the projectile from point of projection up to the top of vertical pole we have
`u_(y) =40 sin theta , a_(yh) =- 9.8 m//s^(2)`,
`t=2 s, y= 20. 4 m`.
As, `y=u_(y) t+1/2 a_(y) t^(2)`
:. 20 .4 =40 sin theta xx 2 1/2 (-9.8) xx 2^(2)`
`=80 sin theta -19.6`
or `sin theta =(20.4 +19.6) // 80 =1//2`
or ` theta =30^(@)`
Horizontal range `=- u^(2)/g sin 2 theta`
`=(40^(2)/(9.8) sin 2 ,xx 30^(@) =141.4 m`.