A) \[12.25\,\,m/s\]
B) \[14.75\,\,m/s\]
C) \[16.23\,\,m/s\]
D) \[17.15\,\,m/s\]
Correct Answer: A
Solution :
Time taken by first stone to reach the water surface from the bridge be t, then \[h=ut+\frac{1}{2}g{{t}^{2}}\Rightarrow 44.1=0\times t+\frac{1}{2}\times 9.8{{t}^{2}}\] \[t=\sqrt{\frac{2\times 44.1}{9.8}}=3\ sec\] 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 \[44.1=u\times 2+\frac{1}{2}9.8{{(2)}^{2}}\] \[\Rightarrow 44.1-19.6=2u\Rightarrow u=12.25\ m/s\]
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