Work Energy Power

JEE Main level, single correct, medium

121 of 183

Question 121 of 183

An object is dropped from a height hh from the ground. Every time it hits the ground it loses 5050% of its kinetic energy. The total distance covered as t→∞t\rightarrow \infty is:

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Answer: (A) 3h3h

Since the object is dropped from height, thus its potential energy is equal to its kinetic energy.
As per question, kinetic energy after hitting the ground is 12mv′2=12×12mv2\dfrac{1}{2}m{v}^{'2}=\dfrac{1}{2}\times \dfrac{1}{2}m{v}^{2}

v′=v2{v}^{'}=\dfrac{v}{\sqrt{2}}

v=euv=eu

e=12e=\dfrac{1}{\sqrt{2}}

H=h(1+e21−e2)H=h\left(\dfrac{1+{e}^{2}}{1-{e}^{2}}\right)

The total distance covered=h(1+121−12)=3h=h\left(\dfrac{1+\dfrac{1}{2}}{1-\dfrac{1}{2}}\right)=3h

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