Question

An object is dropped from height from the ground. Every time it hits the ground it losses $50\% $ of its kinetic energy. The total distance covered as $t \to \infty $ is:
(A) $3h$
(B) $\infty $
(C) $\dfrac{8}{3}h$
(D) $\dfrac{5}{3}h$

Answer 1

Hint It is given that the object is dropped from height from the ground for infinite times. Every time it hits the ground it losses $50\% $ of its kinetic energy. We have to find the distance covered by the object for infinite times, the total distance is the sum of the heights in infinite times. Since the object loses $50\% $ of its kinetic energy every time it hits the ground, its kinetic energy gets decreased by the ratio $\dfrac{1}{2}$, so it forms a G.P sequence. Find the answer by solving the G.P sequence.

Complete step by step answer
Energy of an object or substance is its ability to do work. Energy can be found in different forms. It is clear that energy can neither be created nor be destroyed. But energy can be transformed from one form of energy to another form of energy. Energy can be found in different forms. It is mainly divided into two forms which are kinetic energy and potential energy.
Potential energy is the energy stored or conserved in an object or substance at the state of rest. The energy stored is based on the position, arrangement or state of the object or substance. The potential energy is given by
$P.E = mgh$
Where,
P.E is the potential energy
m is the mass of the object
g is acceleration due to gravity
h is the height from where the object started fall to the ground
The kinetic energy is the energy of an object or substance that it possesses due to its motion. The kinetic energy is given by
$K.E = \dfrac{1}{2}m{v^2}$
Where,
K.E is the kinetic energy
m is the mass of the object
v is the velocity of the object
Free fall is the movement of an object towards gravity where its weight is the only force acting on an object in the presence of gravity.
The object possesses potential energy before starting falling since Potential energy is the energy stored or conserved in an object at the state of rest.
Once it started falling the potential energy of the objects gets converted into kinetic energy since the kinetic energy is the energy of an object or substance that it possesses due to its motion. So in this case potential energy is equal to kinetic energy which is the law of conservation of energy.
$ \Rightarrow mgh = \dfrac{1}{2}m{v^2}{\text{ }} \to {\text{1}}$
Given that the object falls from a height, “h”
It is said that the object losses $50\% $ of its kinetic energy every time it hits the ground
When the object hits the ground for first time the height is ${h_1}$
Then, from above discussion
$ \Rightarrow mg{h_1} = \dfrac{1}{2}\left( {\dfrac{1}{2}m{v^2}} \right)$ Since the object losses $50\% $ of its kinetic energy every time it hits the ground
Substitute equation 1 in above equation
$ \Rightarrow mg{h_1} = \dfrac{1}{2}\left( {mgh} \right)$
$ \Rightarrow {h_1} = \dfrac{1}{2}h$
$ \Rightarrow {h_1} = \dfrac{h}{2}$
Similarly
$ \Rightarrow {h_2} = \dfrac{h}{4}$
$ \Rightarrow {h_3} = \dfrac{h}{8}$
$ \Rightarrow {h_n} = \dfrac{h}{{{2^n}}}$
Then the total distance will be
Total distance covered by the object $ = h + 2{H_1} + 2{h_2} + 2{h_3} + ........$ since $t \to \infty $
$ \Rightarrow {\text{Total distance}} = h + 2{H_1} + 2{h_2} + 2{h_3} + ........$
$ \Rightarrow {\text{Total distance}} = h + 2({h_1} + {h_2} + {h_3} + ........)$
Substitute the value of ${h_1},{h_2},{h_3}.......$
$ \Rightarrow {\text{Total distance}} = h + 2(\dfrac{h}{2} + \dfrac{h}{4} + \dfrac{h}{8} + ........){\text{ }} \to {\text{2}}$
If we take $ \Rightarrow \dfrac{h}{2} + \dfrac{h}{4} + \dfrac{h}{8} + ........$ we can see that this sequence is in G.P progression
For a G.P progression the sum of n terms is given by the formula
${S_n} = \dfrac{{a({r^{n - 1}})}}{{1 - r}}$
For a G.P progression the sum of infinite terms is given by the formula
${S_\infty } = \dfrac{a}{{1 - r}}$
Where,
S is the sum of the terms
a is the first term of the sequence
r is the common ratio by which the sequence is preceding
Now,
$ \Rightarrow \dfrac{h}{2} + \dfrac{h}{4} + \dfrac{h}{8} + ........$ This sequence has infinite number of terms
So, ${S_\infty } = \dfrac{a}{{1 - r}}$
$a = \dfrac{h}{2}$
$r = \dfrac{1}{2}$
\[ \Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}\]
$ \Rightarrow {S_\infty } = \dfrac{{\dfrac{h}{2}}}{{1 - \dfrac{1}{2}}}$
$ \Rightarrow {S_\infty } = \dfrac{{\dfrac{h}{2}}}{{\dfrac{1}{2}}}$
$ \Rightarrow {S_\infty } = h$ i.e. $ \Rightarrow \dfrac{h}{2} + \dfrac{h}{4} + \dfrac{h}{8} + ........ = h$
Substitute this value in equation 2
$ \Rightarrow {\text{Total distance}} = h + 2(\dfrac{h}{2} + \dfrac{h}{4} + \dfrac{h}{8} + ........){\text{ }} \to {\text{2}}$
$ \Rightarrow {\text{Total distance}} = h + 2h$
$ \Rightarrow {\text{Total distance}} = 3h$

Hence the correct answer is option (A) $3h$

Note A G.P sequence is a Geometric progression sequence where each term is the multiplication or division of the preceding term by a common number. We have studied about this in lower classes.