Question

A ball is left to fall from height 10 metre the energy of ball is decreased by 30% after striking the floor up to what height ball will rise after striking ?

Solve it as soon as possible...​

Answer 1

Solution :

Initially ,

• Height of the ball from the ground = 10 m
• Acceleration due to gravity = 10 m/s²

Potential Energy is given by ,

$\implies\mathtt{PE = mgh }$

⇒ PE = 100 m

Now the PE of the ball is decreased by 30  %

Energy left (PE')

= PE - 30 PE/ 100

= 100PE -30 PE / 100

= 70PE / 100

= 70 x 100m / 100

= 70 m

We know that ,

⇒ PE' = mgh'

⇒ 70 m = m x 10  x h'

⇒ h' = 7 m

Hence ,

The will raise to a height of 7 m .

Answer 2

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

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$\large\underline{\green{\bf Given :-}}$

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• Ball is left to fall from height 10 metre

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• The energy of ball is decreased by 30% after striking the floor

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$\large\underline{\red{\bf To \: Find :-}}$

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• Up to what height ball will rise after striking

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$\large\underline{\orange{\bf Solution :-}}$

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• Let "E" be the energy of the ball falling through a height of 10m

• Let "E1" be the energy of the ball when bouncing back

• Let "h1" be the height up to which the ball will rise

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$\underline{\bold{\texttt{Potential energy :}}}$

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➜ E = mgh -------- (1)

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Where ,

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• m = Mass of the body

• g = Gravity

• h = Height

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$\underline{\bold{\texttt{Potential energy at 10 m. :}}}$

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• m = m

• g = 9.8

• h = 10

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⟮ Putting these values in (1) ⟯

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➜ E = mgh

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➜ E = m(9.8)(10)

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➨ E = 98m J

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Now given that the body losses 30% of its energy and rebounds back to a height of “h1" and its new energy is “E1” which is 70% of the initial energy “E”

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➜ E1 = 70% (E) ------- (2)

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$\underline{\bold{\texttt{Potential energy at height "h1" :}}}$

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➜ E1 = mgh1 ------ (3)

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Where,

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• m = Mass of body [Remains same] = m

• g = gravity [Remains same] = 9.8

• h1 = Rebounding height = h1

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⟮ Putting the above values in (3) ⟯

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➜ E1 = mgh1

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➜ E1 = m(9.8)(h1) ------- (4)

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⟮ Putting the value of equation (4) to (2) ⟯

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➜ E1 = 70% (E)

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➜ m(9.8)(h1) = 70% (E)

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➜ $\sf \cancel m(9.8)h1 = \dfrac { \cancel { 70} } { \cancel { 100} } ( 98\cancel m)$

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➜ $\sf 9.8 × h1 = \dfrac { 7 } { 10 } × 98$

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➜ $\sf \cancel { 98} × h1 = \cancel { 98 } × 7$

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➨ h1 = 7

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• Hence the ball will rise upto a height of 7m after striking the ground

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