Question

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

• A. $2s$
• B. $5s$
• C. $7s$
• D. $9s$

Answer 1

The correct option is C $7s$

The component of velocity in the vertical-upward direction is,

$v=50sin\left({30}^0\right)=50\times 0.5=25m/s$

Let the time taken to reach ground from initial position be $tsec$

The acceleration due to gravity is, $g=10m/s^2$, in the vertical-downward direction.

and the distance traveled is, $h=70m$, in the vertical-downward direction.

So here, $h=-vt+\frac{1}{2}g{t}^2⟹70=-25t+(0.5\times 10\times t^2)$

$⟹t^2-5t-14=0$

Solving above quadratic equation, we get $t=-2,7$

As time is always positive, $t=7s$

Option C is correct.