A ball is thrown vertically upwards with a velocity u from the ground the attain a maximum height Hmax then find out the time and displacement at which the ball have half of the maximum speed
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maximum height =(velocity)^2/2×g
half speed =u/2
then maximum height =Hmax/4
and time =half of previous time
half speed =u/2
then maximum height =Hmax/4
and time =half of previous time
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