Question

f the energy of a ball falling from a height of 20 m is reduced by 60 %, how high will it rebound ? *

Answer 1

Answer: 8 m.

Given:

$\rm Height(h)=20\:m$

$\rm Acceleration\:due\:to\:gravity=9.8\: m/s^2$

$\rm Reduced\:energy=60\: \%$

To find:

$\textrm{The height at which the ball rebound.}$

Formula used:

$\boxed{\rm Potential\:Energy(P.E.)=mgh}$

Explanation:

$\textrm{\underline{\textbf{Potential Energy:}} A type of energy which an object possess when it is at rest.}$

$\rm Let\:h'\:be\:the\:height\:at\:which\:the\:ball\:will\:rebound.$

$\textrm{According to the question,}$

$\Rightarrow \rm P.E.\: at \:h=m\times 9.8\times 20$

$\Rightarrow \rm P.E.\:at\:h=196m$

$\textrm{On striking the ground, the ball loses its 60\% energy,}$

$\therefore\textrm{Remaining energy=Total energy-Reduced energy}$

$\Rightarrow \textrm{Remaining energy=100-60}$

$\Rightarrow \textrm{Remaining energy=40\: \%}$

$\Rightarrow \rm Remaining\:energy=\dfrac{40}{100}\times 196m=78.4m$

$\textrm{Hence, the final energy of the ball is,}$

$\Rightarrow \rm mgh'=78.4m$

$\Rightarrow \rm 9.8\times h'=78.4$

$\Rightarrow \boxed{\rm h'=8\:m}$

Hence, the ball will rebound 8 m high.

Answer 2

Answer:

Given That The Ball Is Dropped From A Height,h

1

=10m

Potential Energy At h

1

=mgh

1

Therefore The Potential Energy At h

1

=m×10×10=100mJ

On striking the ground level, the ball looses 40%of its initial energy:

i.e.

100

40

×100mJ=40mJ

The energy left on striking the ground=100mJ−40mJ=60mJ

So the final energy of the ball, mgh

2

=60mJ

i.eh

2

=

10

60

=6m

Hence the ball will bounce back to a height of 6m.