A 500 g stone is thrown up with a velocity of 15 m s⁻¹. Find its (i) P.E. at its maximum height (ii) K.E. when it hits the ground Ans. (56.25 J, 56.25 J)


Given:  

Mass of the stone = m =  500 g = 0.5 kg
Speed of the stone = v = 15 ms⁻¹

To Find:

Potential Energy at maximum height= P.E. = ? J
Kinetic Energy when it hit the ground = K.E. = ? J

Solution:   

According to the law of conservation of enegery, energy can neither be created nor destroyed but only converted from one energy to another. So, in this numerical, the stone thrown vertically contained K.E. totally. Butt this K.E. decrease continuously due to the decrease in speed of the stone and is fully converted into P.E. at maximum height by virtue of its position as it becomes to rest. Similarly, in the vertical downwards direction P.E. decrease continuously because of a decrease in height and is fully converted into K.E. by attaining maximum speed just before hitting the ground. Thus,

P.E. at Maximum Height = K.E. energy Initially when the stone thrown up -------eqn (1)

and

K.E. energy when it hit the ground = P.E. at Maximum Height -------Eqn (2)

As  K.E.  = `\frac {1}{2}`mv²  So, 

P.E. at Maximum Height =  `\frac {1}{2}`mv²

by putting values 

P.E. at Maximum Height =  `\frac {1}{2}`(0.5 kg) (15 ms⁻¹)²

P.E. at Maximum Height =  `\frac {1}{2}`(0.5 kg) (225 m² s⁻²)

P.E. at Maximum Height =  56.25 kg m² s⁻² 

Or

P.E. at Maximum Height =  56.25 J


Now using equation no. (2) that is 

K.E. energy when it hit the ground = P.E. at Maximum Height = 56.25 J   (as calculated above)


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