Question

A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone?

• A. 12.25 m/s
• B. 14.75 m/s
• C. 16.23 m/s
• D. 17.15 m/s

Answer 1

The correct option is A

12.25 m/s

Time taken by first stone to reach the water surface from the bridge be t, then

$h=ut+\frac{1}{2}g{t}^2\Rightarrow 44.1=0\times t+\frac{1}{2}\times 9.8t^2$

$t=\sqrt{\frac{2\times 44.1}{9.8}}=3sec$

Second stone is thrown 1 sec later and both strike simultaneously.This means that the time of travel for the second stone = 3 - 1 = 2 sec

Hence, 44.1 = $2u+\frac{1}{2}9.8(2{)}^2$

$\Rightarrow 44.1-19.6=2u\Rightarrow u=12.25m/s$