Q. A ball is dropped from a bridge at a height of 176.4 m over a river. After 2 s, a second ball is thrown straight downwards. What should be the initial velocity of the second ball so that both hit the water simultaneously?
A) $2.45\ {ms}^{- 1}$
B) $49\ {ms}^{- 1}$
C) $14.5\ {ms}^{- 1}$
D) $24.5\ {ms}^{- 1}$
Explanation
(d) For first ball
$\ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}gt^{2} = 176.4$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ t = \ \sqrt{\frac{176.4\ \times 2}{10}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t = 5.9\ s$
For second ball, $t = 3.9\ s$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u(3.9) + \frac{1}{2}g(3.9)^{2} = 176.4$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.9u + \frac{10}{2}(3.9)^{2} = 176.4$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ u = 25.7\ {ms}^{- 1}$
This value is approximated to $24.5\ {ms}^{- 1}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.