Q. A ball is projected upwards from the top of a tower with a velocity of $50\ ms^{- 1}$ making an angle of $60{^\circ}$ with the vertical. If the height of the tower is 70 m, then the ball will reach the ground in (Take $g = 10\ ms^{- 2}$)
A) 2 s
B) 3 s
C) 5 s
D) 7 s
Explanation
(d) $u = - 50\cos{60{^\circ}}ms^{- 1} = - 25ms^{- 1}$
$a = 10\ ms^{- 2},t = ?,s = 70\ m$
Using $s = ut + \frac{1}{2}at^{2},\ $ we get
$70 = - 25 + 5t^{2}$ or $5t^{2} - 25t - 70 = 0$
or $t^{2} - 5t - 14 = 0$ or $t^{2} - 7t + 2t - 14 = 0$
or $t(t - 7) + 2(t - 7) = 0$ or $(t - 7)(t + 2) = 0$
or $t = 7s$ [Rejecting –ve value of $t$]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.