Question

A body freely falling from rest has a velocity 'v' after it falls through a height 'h' .The distance it has to fall down further for its velocity to become double is:
The Answer is 8h..
I need a solution. ​

Answer 1

Answer:

Use energy conservation twice

From P to Q

$mgh = \frac{1}{2} m {v}^{2} \\ {v}^{2} = 2gh$

From Q to R

$\frac{1}{2} m {v}^{2} + mgx = \frac{1}{2} m(2v) {}^{2} \\ 2 {v}^{2} + 4xg = 8 {v}^{2} \\4gx = 6 {v}^{2} = 6 \times 2gh \\ x = 3h$

I am 100% sure it is right answer

Answer 2

Answer:

just solve it to get answer ( it seems answer is Wrong )