A ball is projected upwards from the top of a tower with a velocity `50ms^-1` making an angle `30^@` with the horizontal. The height of tower is 70m. After how many seconds from the instant of throwing, will the ball reach the ground. `(g=10 ms^-2)`

Options

A. `2s`
B. `5s`
C. `7s`
D. `9s`

Explanation

(c) Taking vertical downward motion of projection from point of projection to grond we have
`u=-50sin 30^(@)=-25m//`
`a=+10m//s^(2),s=70m,t=?`
`s=ut+1/2at^(2)`
So, `70=-25xxt+1/2xx10xxt^(2)`
or `5t^(2)-25t-70=0` or `t^(20-5t-14=0`
On solving `=7s`