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 u2 2 1 2 : u ( Assume upward acceleration is –g and downward acceleration to be +g ).

Answer 1

Answer:

2=u2+2as

0=(u1)2+2(−g)s1

Hence, s1=u12/2g

Similarly, s2=u22/2g

Therefore, s2s1=u22u12

or

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