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 ?

(1) 12.25 m/s

(2) 14.75 m/s

(3) 16.23 m/s

(4) 17.15 m/s

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Concept Questions :-

Uniformly accelerated motion

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

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

$t=\sqrt{\frac{2×44.1}{9.8}}=3\text{\hspace{0.17em}}sec$

Second stone is thrown 1 sec later and both strikes simultaneously. This means that the time left for second stone $=3-1=2\text{\hspace{0.17em}}sec$

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

$⇒44.1-19.6=2u⇒u=12.25\text{\hspace{0.17em}}m/s$

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