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 to become triple

Answer 1

Answer:

√3h

Explanation:

${v}^{2} = {u}^{2} + 2as$

where,

• v=final velocity
• u=initial velocity=0 (rest)
• a=acceleration=g (freefall)
• s=displacement=h (height)

therefore,

${v}^{2} = {0}^{2} + 2gh \\ {v}^{2} = 2gh \\ v = \sqrt{2gh}$

from this equation,

$v \alpha \sqrt{h}$

hence, for velocity to become triple,height should be√3h

$3v \alpha \sqrt{3h}$

hope it helps...!!