Question

A ball is dropped from the top of a tall tower. The
ratio of change in kinetic energy of the ball in first
and third second of its motion will be
(1) 1:5
(2) 1:9
(3) 1:3
(4) 1:8​

Answer 1

Answer:

a) 1:5,,.....,...,.,...,...,......,.,

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Answer 2

The ratio of change in kinetic energy of the ball in first  and third second of its motion is 1:5.

(1) is correct option.

Explanation:

Given that,

A ball is dropped from the top of a tall tower.

We need to calculate the velocity

Using equation of motion

$v=u-gt$

For initial, u = 0

Put the value in the equation

$v_{t}=-gt$

Now, the velocity of the ball at first second

$v_{1}=-g$

The velocity of the ball at third second

$v_{3}=v_{3}-v_{2}$

We need to calculate the ratio of change in kinetic energy of the ball in first and third second of its motion

Using formula of kinetic energy

$\dfrac{K.E_{1}}{K.E_{3}}=\dfrac{\dfrac{1}{2}mv_{1}^2}{\dfrac{1}{2}mv_{3}^2}$

$\dfrac{K.E_{1}}{K.E_{3}}=\dfrac{v_{1}^2}{v_{3}^2}$

$\dfrac{K.E_{1}}{K.E_{3}}=\dfrac{g^2}{v_{3}^2-v_{2}^2}$

$\dfrac{K.E_{1}}{K.E_{3}}=\dfrac{g^2}{5g^2}$

$\dfrac{K.E_{1}}{K.E_{3}}=\dfrac{1}{5}$

Hence, The ratio of change in kinetic energy of the ball in first  and third second of its motion is 1:5.

Learn more :

Topic : kinetic energy

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