Question

A ball is projected upwards from the top of the tower with a velocity $50{\text{ms}}^{-1}$ making an angle of ${30}^{\circ }$ with the horizontal. The height of tower is $70\text{m}$. After how many seconds the ball will strike the ground?

• A. $3\text{sec}$
• B. $5\text{sec}$
• C. $7\text{sec}$
• D. $9\text{sec}$

Answer 1

The correct option is C $7\text{sec}$

By the third equation of motion, $($let downward direction be $+ve)$,
$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$a_y=10{\text{m/s}}^2,s_y=70\text{m}$ and $t$ is the time taken for the ball to reach the ground
$70=-\left(50\sin {30}^{\circ }\right)t+\frac{1}{2}\times 10\times t^2$
$70=-25t+5t^2$

$t^2-5t-14=0\Rightarrow t=-2\text{sec}$ $(-ve$ time is not possible$)$

or $t=7\text{sec}$