Q. A ball is projected upwards from the top of tower with a velocity $50\ ms^{- 1}$ making an angle $30{^\circ}$ with the horizontal. The height of tower is 70m. After how many seconds from the instant of throwing will the ball reach the ground?
A) 2 s
B) 5 s
C) 7 s
D) 9 s
Explanation
(c) Taking vertical downward motion of projectile from point of projection to ground, we have
$u = - 50\sin{30{^\circ}} = - 25\ ms^{- 1}$
$a = + 10\ ms^{- 2},s = 70\ m,\ t = ?$
$\therefore\ \ s = ut + \frac{1}{2}at^{2};$
So, $70 = - 25 \times t + \frac{1}{2} \times 10 \times t^{2}$
or $5t^{2} - 25t - 70 = 0$ or $t^{2} - 5t - 14 = 0$
On solving $t = 7\ s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.