Two balls are dropped from the same height at two different places A and B where the acceleration due to gravities 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 $υ$ m/s. Then the value of ‘ $υ$ /t’ is

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$2 \sqrt{g_{A} g_{B}}$

$\frac{1}{g_{A} g_{B}}$

$\sqrt{g_{A} g_{B}}$

$\frac{1}{\sqrt{g_{A} g_{B}}}$

answer is D.

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Detailed Solution

s → distance covered by each car time taken to complete journey by two cars t₁ and t₂ and velocities c₁ and c₂.

According to given problem
v₁ – v₂ = v and t₂ – t₁ = t
now $\frac{v}{t} = \frac{v_{1} - v_{2}}{t_{1} - t_{2}} = \frac{\sqrt{2 g_{A} s} - \sqrt{2 g_{B} s}}{\sqrt{\frac{2 s}{g_{B}}} - \sqrt{\frac{2 s}{g_{A}}}}$

$\begin{array}{l} = \frac{\sqrt{g_{A}} - \sqrt{g_{B}}}{\sqrt{\frac{1}{g_{A}}} - \sqrt{\frac{1}{g_{B}}}} \\ \frac{v}{t} = \sqrt{g_{A} g_{B}} \end{array}$