1.

A man is throwing balls in air. He throws next ball when previous one is at the highest point. If he throws each ball after 2s , then height to which ball rises is

Answer»

10 m
20 m
30 m
15 m

Solution :The ball reaches the highest points in 2 s . Now
v= u+at `:.0=u-gxx2`
or `u=2xx10=20 ms^(-1)`
ALSO `h=(v^(2)-u^2)/(2a)=(0-400)/((-10)xx2)=20 m`


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