A body of mass 2.0 kg is dropped from a rest position 5m above ground. What is its velocity at a height of 3.0 m above the ground? (v = 6.3 m s⁻¹)


Given:

Mass of the body = m = 70 kg
Height from the ground = h1 = 5 m 
Height at velocity is to be found = h2 = 5 m 
Effective height through which the body fall = h = h1 - h2 = 5 m - 3 m = 2 m
Value of g = 9.81 m s⁻²

To Find:

The velocity of the body = v =?


Solution:

According to the law of conservation of energy

Gain in K.E = Loss in P.E.

`\ frac{1}{2}`mv² = mgh

v² = 2gh

v = `\ sqrt {2gh}`

by putting values

v = `\ sqrt {2ｘ9.81 m s⁻²ｘ 2 m}`

v = `\ sqrt {39.24 m² s⁻²}`

v = 6.264 m s⁻¹

or

v = 6.3 m s⁻¹ ----------------Ans



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