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 strikes simulataniously. This
Means that the time left for second stone =3-1=2 sec

Hence $44.1-u\times 2+\frac{1}{2}9.8(2{)}^2$
$\Rightarrow$ 44.1-19.6=2u$\Rightarrow$ u=12.25 m/s