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 $70\ m$. 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) The vertical component of velocity of projection
$= - 50\sin{30{^\circ} = - 25\ m/s}$
If $t$ be the time taken to reach the ground,
$h = ut + \frac{1}{2}gt^{2} \Rightarrow 70 = - 25t + \frac{1}{2} \times 10t^{2}$
$\Rightarrow 70 = - 25t + 5t^{2} \Rightarrow t^{2} - 5t - 14 = 0 \Rightarrow t = - 2s$ and $7s$
Since, $t = - 2\ s$ is not valid $\therefore t = 7s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.