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 /2_g 

For first ball, h₁ = u² / 2_g 

and for second ball, h₂= u² / 2_g