Kinematics Question 180
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
Options:
A) $ 12.25m/s $
B) $ 14.75m/s $
C) $ 16.23m/s $
D) $ 17.15m/s $
Show Answer
Answer:
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}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\ 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 $
BETA
