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)

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 $