Question

A body is dropped from a height h. By using the concept that during the fall the body's total mechanical energy remains constant, find its speed when the body is at height h/4. (ignoring air resistance).

Answer 1

Answer: The speed is $v = \sqrt{\dfrac{3}{2}gh}$.

Explanation:

We know that mechanical energy or total energy is the sum of potential energy and kinetic energy and according to law of conservation of energy the total mechanical energy remains constant

$mgh + 0 = mgh + \dfrac{1}{2}mv^{2}$

$gh - \dfrac{gh}{4}= \dfrac{1}{2}v^{2}$

$v^{2} = \dfrac{3}{2}gh$

$v = \sqrt{\dfrac{3}{2}gh}$

Hence, the speed is $v = \sqrt{\dfrac{3}{2}gh}$.

Answer 2

Total energy of the body at height = mgh. (All potential energy)

Total energy of the body on ground = mv^2/2 (all Kinetic energy).

Total energy at height h/4 = Potential energy at that height + Kinetic Energy at that height

= mgh/4 + mv^2/2  = mgh (Total energy is always constant)

mv^2/2 = 3mgh/4

v^2 = 3gh/2

v = √(3gh/2)