Question

A ball is dropped from a height of 10m. It strikes the ground and rebounds up to a height of 2.5m. During the collision the percentage loss in the kinetic energy is:

Answer 1

K.E=(mv^2)/2

=>v1 = root(2gh)

=>v1 = root(200)

=>v1 = 14.1 m/s => K.E = m(196)/2 = 98m

Now,

=>v2 =root(2gh)

=root(50)

= 7.01m/s => K.E = m(50)/2 = 25m

Difference = 98m-25m = 73m

Percentage of loss:

$\frac{73m}{98m} \times 100$
=74.1%

PLEASE NOTE MY ANSWER AS BRAINLIEST
EVEN MY ANSWER IS WRONG

Answer 2

Answer:

The percentage loss in the kinetic energy is 75%.

Explanation:

Given that,

Height h = 10 m

Using third equation of motion,

$v_{1}^2=u^2+2gs$

$v_{1}^2=2gh$

$v_{1}=\sqrt{2gh}$

Put the value of h and g

$v_{1}=\sqrt{2\times9.8\times10}$

$v_{1}=14\ m/s$

The kinetic energy is

$K.E=\dfrac{1}{2}mv^2$

The initial kinetic energy

$K.E_{i}=\dfrac{1}{2}\times m\times (14)^2$

$K.E_{i} = 98m$

After rebounds up to a height of 2.5 m.

Using equation of motion again

$v_{2}^2=0+2gh$

$v_{2}=\sqrt{2\times9.8\times2.5}$

$v_{2}=7\ m$

The final kinetic energy is

$K.E=\dfrac{1}{2}mv^2$

$K.E_{f}=\dfrac{1}{2}\times m\times (7)^2$

$K.E_{f} = 24.5m$

The percentage loss in the kinetic energy is

$loss\ of\ K.E=\dfrac{K.E_{i}-K.E_{f}}{K.E_{i}}\times100$

$loss\ of\ K.E=\dfrac{98-24.5}{98}$

The percentage Loss of kinetic energy = 75%

Hence, The percentage loss in the kinetic energy is 75%.