Question

A body freely falling from the rest has a velocity v after his falls through a height h. The distance, it has to fall down further for its velocity becomes double, is

• A. $4\text{ }h$
• B. $2\text{ }s$
• C. $1/\sqrt{2}s$
• D. $\sqrt{2}s$

Answer 1

Answer: (A)

$4\text{ }h$

Answer 2

Here, initial velocity of the body ${{v}_{1}}=v$ Initial height $h=hc$ Final velocity of the body ${{v}^{2}}=2v$ Now, from the equation of motion ${{v}^{2}}={{u}^{2}}+2gh$ ${{v}^{2}}\propto h$ So, ${{\left[ \frac{{{v}_{1}}}{{{v}_{2}}} \right]}^{2}}=\left[ \frac{{{h}_{1}}}{{{h}_{2}}} \right]$ Or, ${{\left[ \frac{v}{2v} \right]}^{2}}=\frac{h}{{{h}_{2}}}$ Or, $\frac{h}{{{h}_{2}}}=\frac{1}{4}$ Or, ${{h}_{2}}=4h$