A) 120
B) 80
C) 60
D) 40
Correct Answer: C
Solution :
Maximum height of ball = 5 m So velocity of projection \[\Rightarrow \]\[u=\sqrt{2gh}\]\[=10\ m/s\] Time interval between two balls (time of ascent) \[=\frac{u}{g}=1\ sec=\frac{1}{60}min\]. So number of ball thrown per min. = 60You need to login to perform this action.
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