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Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them 

would be in the ratio of u 2/t ; u 2/2 ( Assume upward acceleration is –g and downward acceleration to be +g ).

1 Answer

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We know for upward motion, v2 = u2 – 2 g h or h =  u2 -v2 /2g 

But at highest point v = 0

Therefore, h =  u2 /2

For first ball, h1 = u2 / 2

and for second ball, h2= u2 / 2g

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