Question

The velocity acquired by a body when it falls through $a$ height $h$ is $v$. If it further falls through a height $x$ $(x<<h)$ , the increase in velocity is approximately

• A. $\frac{vx}{2h}$
• B. $\frac{2hv}{x}$
• C. $\frac{vx}{h}$
• D. $\frac{vh}{x}$

Answer 1

Answer: (A)

$\frac{vx}{2h}$

$V^2=u^2+2gh$ - (1)
$V_1^2=u^2+2g(h+x)$
increase in velocity $\Rightarrow V_1-V.$
$\Rightarrow V_1-V=\sqrt{2g(h+x)}-\sqrt{2gh}$
since $x<<<<h.$
$\Rightarrow \sqrt{h}\left(2g\right(1+\frac{x}{2h})-2g)$
$\Rightarrow \frac{gx}{\sqrt{h}}=\Delta V$
Since $V=\sqrt{2gh}\Rightarrow g=\frac{V}{\sqrt{2}h}$
$\Rightarrow \Delta V=\frac{Vx}{2h}$