Question

A ball is thrown up is caught back by thrower after 6sec.Calculate the velocity with the ball was thrown up.The maximum height attained by the ball.The distance of the ball below the highest point after 2sec.Take g=10m/s square​

Answer 1

Answer:

wkt 6=2(2h/g)^1/2

so h=45m(maximum height)

velocity with which it was thrown up=(2gh)^1/2=30m/s

on it's way back to the thrower's hand we can assume the initial velocity of the ball to be zero since it is starting from rest at the highest position.

so the required distance=1/2gt^2=20m

hope it helps

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