Question

Two balls are dropped from the same height at
two different places A and B where the
acceleration due to gravity are gA and gs. The
body at 'B' takes 'seconds less to reach the
ground and striks the ground with a velocity
greater than at 'A' by vm/s. Then the valve of
v/t is
​

Answer 1

Complete Question:

Two balls are dropped from the same height at two different places A and B where the acceleration due to gravity are $g_A$ and $g_B$. The body at ‘B’ takes ‘t’ seconds less to reach the ground and strikes the ground with a velocity greater than at ‘A’ by v m/s . Then the value of v/t is

Given:

• Acceleration due to gravity at A =  $g_A$

• Acceleration due to gravity at B =  $g_B$

• The  body at 'B' takes 't' seconds less to reach the  ground and strikes the ground with a velocity  greater than at 'A' by v m/s.
• Height is same , h.

To find:

• Value of v/t

Answer:

• t = time taken by A - time taken by B

          = $\sqrt{\frac{2h}{g_A}} -\sqrt{\frac{2h}{g_B}}$       ---------Equation 1

• v = Velocity at B - Velocity of A

           = $\sqrt{2g_A h} -\sqrt{2g_B h}$    ----------Equation 2

• Dividing equation 1 by 2, we get: $\frac{v}{t} = \sqrt{g_A g_B}$

• Hence, value of (v/t) is $\sqrt{g_A g_B}$