Question

An object is dropped from a height of 20m.Using law of conservation of energy,find its speed when it's about to hit the ground?
class 9​

Answer 1

Given:

Object is dropped from 20 m height.

To find:

Speed of body when it strikes ground (using law of conservation of energy) ?

Calculation:

• The Law of Conservation of Energy states that the total amount of mechanical energy of a freely falling body will always remain constant.

$PE1 + KE1 = PE2 + KE2$

$\implies \: mgh + 0 = 0 + \dfrac{1}{2} m {v}^{2}$

$\implies \: m \times 10 \times 20 = \dfrac{1}{2} m {v}^{2}$

$\implies \: 10 \times 20 = \dfrac{1}{2} {v}^{2}$

$\implies \: v = \sqrt{400}$

$\implies \: v = 20 \: m {s}^{ - 1}$

So, velocity at ground is 20 m/s.