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.
• B.
• C.
• D.

Answer 1

Answer: (A)

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 simultaneously. This means that the time left for second stone = 3 - 1 = 2 sec
Hence $44.1=u\times 2+\frac{1}{2}9.8(2{)}^2$
$\Rightarrow 44.1-19.6=2u\Rightarrow u=12.25\frac{m}{s}$