Question

A ball is projected upwards from the top of a tower with a velocity $50\text{m/s}$ making an elevation angle ${30}^{\circ }$ with the horizontal. The height of the tower is $70\text{m}$. Find time taken by the ball to reach the ground from the instant of throwing.

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

Answer 1

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


Velocity component along $\text{y}-$axis
$U_y=50\sin {30}^{\circ }=25\text{m/s}$ (vertically upward direction)

Let, the time taken to reach ground from initial position be $t\text{sec}$.

The acceleration due to gravity is, $g=10{\text{m/s}}^2$ , in the vertical-downward direction.
And the distance travelled is, $h=70\text{m}$, in the vertical-downward direction.

So, using the equation of motion,

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

we get,

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

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

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

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

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

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

$\therefore t=-2$ & $t=7$

Since, $t=-2$ is not a valid data. So,
$t=7\text{sec}$.

Hence, option (c) is the correct answer.