Question

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)`.

Answer 1

Let ` theta`be the angle of projection of projectile with horizontal. Taking vertical upward motion of the projectile from point projection up to the top of vertical pole, we have
` u_y = u sin theta = 40 sin theta, a_y =- 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 0 xx 2^2`
or ` is theta = ( 20 . 4 + 19 .6) //80 = 1//2 or theta = 30^2`
Horizontal range, ` R = (u^2 sin 2 theta) /g`
`= ( 40^2)/( 9.8) sin ( 2 xx 30^@) = 141.4 m`.