Question

A body freely falling from rest has a velocity V after it falls through a height h. The distance it has to fall further for its velocity to be come double is : (A) 3 h (B) 6 h (C) 8 h (D) 10 h

Answer 1

When the body falls through a height h from rest,

• displacement, s = h

• initial velocity, u = 0

• final velocity, v = V

• acceleration, a = g

By third equation of motion,

$\longrightarrow\sf{v^2=u^2+2as}$

$\longrightarrow\sf{V^2=2gh\quad\quad\dots(1)}$

Let the body fall further by a distance h', after which the velocity becomes 2V.

• Total displacement when fallen from rest, s = h + h'

• Initial velocity, u = 0

• Final velocity, v = 2V

• Acceleration, a = g

By third equation of motion,

$\longrightarrow\sf{v^2=u^2+2as}$

$\longrightarrow\sf{(2V)^2=2g(h+h')}$

$\longrightarrow\sf{4V^2=2gh+2gh'}$

From (1),

$\longrightarrow\sf{4\cdot2gh=2gh+2gh'}$

$\longrightarrow\sf{8gh=2gh+2gh'}$

$\longrightarrow\sf{2gh'=6gh}$

$\longrightarrow\underline{\underline{\sf{h'=3h}}}$

Hence (A) is the answer.

Answer 2

Answer:

(a) 8h

Explanation: