A)
B)
C)
D)
Correct Answer: A
Solution :
The kinematical equations of motion of the ball while falling down is, \[V=-\sqrt{2gh}\] So, speed just before the impact, \[{{V}_{1}}=-\sqrt{2gd}\] and speed just after the impact, \[{{V}_{2}}=\sqrt{2g(d/2)}\] Afterword?s \[V=\sqrt{V_{2}^{2}-2gh}\]You need to login to perform this action.
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