Question

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 ).

Answer 1

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

But at highest point v = 0

Therefore, h = u2 /2g

For first ball, h1 = u2 / 2g

and for second ball, h2= u2 / 2g