Question

A ball is projected upwards from the top of a tower with a velocity $50\text{m/s}$ making an angle ${30}^{\circ }$ with the horizontal. The height of the tower is $70\text{meter}$. After how much time from the instant of throwing, the ball will reach the ground?

• A. $2\text{sec}$
• B. $5\text{sec}$
• C. $7\text{sec}$
• D. $9\text{sec}$

Answer 1

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


From above diagram,

$u_x=50\cos {30}^{\circ }=25\sqrt{3}\text{m/s}$, and

$u_y=50\sin {30}^{\circ }=25\text{m/s}$ (upward)

In order to reach ground it towards $70\text{m}$ downward in $y$ direction under the effect of gravity.

Applying <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> 2nd Equation of Motion

$h=u_{y}t+\frac{1}{2}{a}_y{t}^2$

here,
$h=-70\text{m};u_y=25\text{m/s};a_y=-g$

$\Rightarrow -70=25t-\frac{1}{2}g{t}^2$

$\Rightarrow -70=25t-5t^2\Rightarrow -14=5t-t^2$

$\Rightarrow t^2-5t-14=0\Rightarrow t^2-7t+2t-14=0$

$\Rightarrow t(t-7)+2(t-7)=0$

$\therefore t=-2\text{s}\text{or}t=7\text{s}$

Since, $t=-2$ isn't valid.

So, $t=7\text{sec}$

Hence, option $\left(c\right)$ is correct.