Question

A stone is dropped into water from a bridge $44.1m$ 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. $12.25m/s$
• B. $14.75m/s$
• C. $16.23m/s$
• D. $17.15m/s$

Answer 1

Answer: (A)

$12.25m/s$

Answer 2

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

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

$t = \sqrt{\frac{2 \times 44.1}{9.8}} = 3\mspace{6mu}\sec$

Second stone is thrown 1 sec later and both strikes simultaneously. This means that the time left for second stone $= 3 - 1 = 2\mspace{6mu}\sec$

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

$\Rightarrow 44.1 - 19.6 = 2u \Rightarrow u = 12.25\mspace{6mu} m/s$