First slide
Projection Under uniform Acceleration
Question

A body of mass m thrown horizontally with velocity v, from the top of tower of height h touches the level ground at a distance of 250 m from the foot of the tower. A body of mass 2 m thrown horizontally with velocity v2, from the top of tower of height 4h will touch the level ground at a distance x from the foot of tower. The value of x is

Moderate
Solution

t = 2hg , x = vt = v2hg

x' = v22(4h)g  x = 250 m

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