First, we have to find the potential energy :
[tex]\bigstar[/tex] A ball at rest is dropped from a height of 12 m.
Given :
According to the question by using the formula we get,
[tex]\implies \sf\boxed{\bold{P.E. =\: mgh}}\\[/tex]
where,
So, by putting those values we get,
[tex]\implies \sf P.E. =\: m \times 10 \times 12\\[/tex]
[tex]\implies \sf P.E. =\: m \times 120\\[/tex]
[tex]\implies \sf\bold{P.E. =\: 120m}\\[/tex]
Hence, the potential energy is 120m .
Now,
[tex]\bigstar[/tex] The final kinetic energy will be equal to initial potential energy.
So,
[tex]\implies \sf\boxed{\bold{K.E. =\: \dfrac{1}{2}mv^2}}\\[/tex]
As,
[tex]\small \implies \sf\bold{Potential\: Energy =\: Kinetic\: Energy}\\[/tex]
[tex]\implies \sf\bold{120m =\: \dfrac{1}{2}mv^2}\\[/tex]
Again, now we have to find the energy lost :
[tex]\bigstar[/tex] It looses 25% of its kinetic energy on striking the ground.
[tex]\small \leadsto \sf\bold{Energy\: Lost =\: 25\%\: of\: Kinetic\: Energy}\\[/tex]
[tex]\leadsto \sf Energy\: Lost =\: \dfrac{25}{100} \times 120m\\[/tex]
[tex]\leadsto \sf Energy\: Lost =\: \dfrac{3000m}{100}\\[/tex]
[tex]\leadsto \sf\bold{Energy\: Lost =\: 30m}\\[/tex]
Hence, the energy lost is 30m .
So, the remaining energy will be :
[tex]\mapsto \sf Remaining\: Energy =\: 120m - 30m\\[/tex]
[tex]\mapsto \sf\bold{Remaining\: Energy =\: 90m}\\[/tex]
Hence, the remaining energy is 90m .
Now, we have to find the height it will bounce :
[tex]\implies \sf 90m =\: m \times 10 \times h\\[/tex]
[tex]\implies \sf 90m =\: 10m \times h\\[/tex]
[tex]\implies \sf \dfrac{90\cancel{m}}{10\cancel{m}} =\: h\\[/tex]
[tex]\implies \sf \dfrac{9\cancel{0}}{1\cancel{0}} =\: h\\[/tex]
[tex]\implies \sf 9 =\: h\\[/tex]
[tex]\implies \sf\bold{\underline{h =\: 9m}}\\[/tex]
[tex]\small \sf\boxed{\bold{\therefore\: The\: height\: it\: will\: bounce\: is\: 9m\: .}}\\[/tex]
Refer to the attachment
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Verified answer
Given :-
To Find :-
Solution :-
First, we have to find the potential energy :
[tex]\bigstar[/tex] A ball at rest is dropped from a height of 12 m.
Given :
According to the question by using the formula we get,
[tex]\implies \sf\boxed{\bold{P.E. =\: mgh}}\\[/tex]
where,
So, by putting those values we get,
[tex]\implies \sf P.E. =\: m \times 10 \times 12\\[/tex]
[tex]\implies \sf P.E. =\: m \times 120\\[/tex]
[tex]\implies \sf\bold{P.E. =\: 120m}\\[/tex]
Hence, the potential energy is 120m .
Now,
[tex]\bigstar[/tex] The final kinetic energy will be equal to initial potential energy.
So,
Given :
According to the question by using the formula we get,
[tex]\implies \sf\boxed{\bold{K.E. =\: \dfrac{1}{2}mv^2}}\\[/tex]
where,
As,
[tex]\small \implies \sf\bold{Potential\: Energy =\: Kinetic\: Energy}\\[/tex]
So,
[tex]\implies \sf\bold{120m =\: \dfrac{1}{2}mv^2}\\[/tex]
Again, now we have to find the energy lost :
[tex]\bigstar[/tex] It looses 25% of its kinetic energy on striking the ground.
So,
[tex]\small \leadsto \sf\bold{Energy\: Lost =\: 25\%\: of\: Kinetic\: Energy}\\[/tex]
[tex]\leadsto \sf Energy\: Lost =\: \dfrac{25}{100} \times 120m\\[/tex]
[tex]\leadsto \sf Energy\: Lost =\: \dfrac{3000m}{100}\\[/tex]
[tex]\leadsto \sf\bold{Energy\: Lost =\: 30m}\\[/tex]
Hence, the energy lost is 30m .
So, the remaining energy will be :
[tex]\mapsto \sf Remaining\: Energy =\: 120m - 30m\\[/tex]
[tex]\mapsto \sf\bold{Remaining\: Energy =\: 90m}\\[/tex]
Hence, the remaining energy is 90m .
Now, we have to find the height it will bounce :
Given :
According to the question by using the formula we get,
[tex]\implies \sf\boxed{\bold{P.E. =\: mgh}}\\[/tex]
[tex]\implies \sf 90m =\: m \times 10 \times h\\[/tex]
[tex]\implies \sf 90m =\: 10m \times h\\[/tex]
[tex]\implies \sf \dfrac{90\cancel{m}}{10\cancel{m}} =\: h\\[/tex]
[tex]\implies \sf \dfrac{9\cancel{0}}{1\cancel{0}} =\: h\\[/tex]
[tex]\implies \sf 9 =\: h\\[/tex]
[tex]\implies \sf\bold{\underline{h =\: 9m}}\\[/tex]
[tex]\small \sf\boxed{\bold{\therefore\: The\: height\: it\: will\: bounce\: is\: 9m\: .}}\\[/tex]
Refer to the attachment