Question

A ball is dropped from rest from a height of 12 m if the ball loses 25% of its kinetic energy on striking the ground how height will it bounce

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Answer 1

Answer:

Before the ball is dropped, Potential Energy stored in the ball is mgh.

According to the question, 'h' = 12 m and 'g' is 10 m/s², 'm' is 'm'.

Potential Energy Stored = m × 10 × 12 = 120.m

By Conservation of Energy, we can say that Final Kinetic Energy while striking is equal to the initial Potential Energy.

⇒ 1/2 mv² = 120.m

It is given that on striking the ground, the ball loses 25% of Kinetic Energy.

⇒ Energy Lost = ( 25/100 ) × 120.m = 0.25 × 120.m = 30.m

Therefore Energy lost is 30.m

Hence Energy still remaining is 120.m - 30.m = 90.m

With this energy the ball rises to a height 'h'. Therefore Potential Difference at that height is equal to 90.m. Therefore on applying the formula we get,

⇒ m × 10 × h = 90.m

⇒ h = 90.m / ( 10.m )

⇒ h = 9

Therefore the height it will bounce is 9m.

Answer 2

$\bold{\underline{\underline{Answer:}}}$

$\bold{\underline{\underline{Step\:by\:step\:explanation:}}}$

I will be using g = 9.8 m/s².

Kindly refer to @KalpeshPrabhakar's answer for calculation with g = 10 m/s²

Given :

• A ball is dropped from rest from a height of 12 m
• Ball loses 25% of its kinetic energy on striking the ground

To find :

• Height it will bounce after striking the ground.

Solution :

The ball at rest is dropped from a height of 12 m.

This means the ball stores potential energy.

We know the formula for Potential Energy.

$\bold{\boxed{\blue{\rm{Potential\:Energy\:=\:mgh}}}}$

We know two quantities out of the three in the formula and the one is unknown.

g = 9.8 m/s²

h = 12 m

Block in the values,

$\rightarrow$$\rightarrow$ $\bold{P.E\:=\:m\times\:9.8\times\:12}$

$\rightarrow$$\rightarrow$ $\bold{P.E=\:117.6m}$

•°• Potential energy = 118 m

[Approx]

Using principle of :-

• Conversation of Energy

The final kinetic energy at the time of striking will be equal to the initial potential energy.

$\rightarrow$$\bold{\boxed{\blue{\rm{Kinetic\:Energy\:={\dfrac{1}{2}mv^2}}}}}$

Potential energy = Kinetic Energy

$\rightarrow$$\bold{118.m={\dfrac{1}{2}mv^2}}$

Energy lost :

The ball loses 25% of its kinetic energy.

$\rightarrow$$\bold{\dfrac{25}{100}}\times{118.m}$

$\rightarrow$$\bold{0.25\times\:118.m}$

$\rightarrow$$\rightarrow$ $\bold{29.5.m}$

Energy lost at the time of striking is 29.5•m

Now the energy remaining in the ball will be difference of potential energy and energy lost while striking.

$\rightarrow$$\bold{Enery\:remaining\:=\:118.m-29.5.m}$

$\rightarrow$$\bold{Energy\:remaining\:=\:88.5.m}$

•°• The ball still posses 88.5• m energy. With this remaining energy the ball bounces to height, h.

Using the formula for potential energy.

P. E = mgh

$\rightarrow$$\bold{88.m}$ $\bold{m\times\:9.8\times\:h}$

$\rightarrow$$\bold{88.m}$ $\bold{9.8.m\times\:h}$

$\rightarrow$$\bold{\dfrac{88.5.m}{9.8.m}}$ = h

$\rightarrow$$\bold{9.039}$ = h

•°• Height the ball will bounce is 9.039 m.