A ball is thrown at an angle of `30^(@)` with respect to the horizontal from the top of a tower 100m high with a speed of `40ms^(-1)`. Calculate the time taken by the ball to reach the foot of the tower. With what velocity will the ball hit the ground ? `(g=10ms^(-2))`

Explanation

Given `theta=30^(@),h=100m, v_0=40ms^(-1)`
`g=10ms^(-2),y=-100` (displacement)
By applying
`y=(v_0sintheta)t-1/2gt^2`
We write
`-100=40sin30^(@)t-5t^2`
`-100=40xx1//2t-5t^2`
i.e `t^2-4t-20=0`
`t=(4+-sqrt(16+80))/2`
i.e t=6.98
By applying `v_y=v_0sintheta-"gt"` we ge
`v_y=40sin30^(@)-10(6.9)`
i.e `v_y=40xx0.5-6.9`
or `v_y=20-69=-49ms^(-1)`
`v_y=-49ms^(-1)`
Negative sign indicates that the velocity is opposite to the initial direction of velocity
`:.` Velocity wiht which the ball strikes the ground `=49ms^(-1)`