Question

Two balls are dropped from the same height from places $A$ and $B$. The body at $B$ takes two seconds less to reach the ground at $B$ strikes the ground with a velocity greater than at $A$ by $10m/s$. The product of the acceleration due to gravity at the two places $A$ and $B$ is:

• A. $5$
• B. $25$
• C. $125$
• D. $12.5$

Answer 1

The correct option is B $25$

Given, $V_A+10=V_B$

$V_A-V_B=-10$

$V_B-V_A=10⟶\left(1\right)$

Similarly, $t_A-t_B=2⟶\left(2\right)$

$\left(1\right)÷\left(2\right)$

$\frac{10}{2}=\frac{V_B-V_A}{t_A-t_B}$

$\Rightarrow 5=\frac{\sqrt{2g_{B}h}-\sqrt{2g_{A}h}}{\sqrt{\frac{2h}{g_A}}-\sqrt{\frac{2h}{g_B}}}$

$\Rightarrow 5=\frac{\sqrt{2h}(\sqrt{g_B}-\sqrt{g_A})}{\sqrt{2h}(\frac{1}{\sqrt{g_A}}-\frac{1}{\sqrt{g_B}})}$

$\Rightarrow 5=\sqrt{g_A{g}_B}$

$g_A{g}_B=25$

So, product of acceleration of gravity of two place.$A$ and $B$ $125$.