Question

* (7) If the energy of a ball falling from a
height of 10 metres is reduced by 40%, how high
will it rebound?​

Answer 1

Given :-

▪ Energy of a ball falling from a height of h = 10 m is reduced by 40% when it strikes the surface.

To Find :-

▪ How high will it rebound.

Solution :-

Let the mass of the ball be m. So, when it was at a height of h = 10 m, It had no kinetic energy but only potential energy.

⇒ Potential Energy = mgh

⇒ P.E = m × 10 × 10 [ g = 10 m/s² ]

⇒ P.E = 100m J

Now, When it stroke the ground it had no potential energy but only kinetic energy which is equal to 100m J (According to the conservation of energy)

Now, It is given that the energy of the ball was reduced by 40% when it stroke the ground, So after striking the ground it had ,

⇒ k.e = K.E - 40% of K.E

⇒ k.e = mgh - 40mgh / 100 [ K.E = mgh ]

⇒ k.e = 100m - 40×100m/100

⇒ k.e = 100m - 40m

⇒ k.e = 60m J

Now, According to the conservation of energy, The kinetic energy will be converted to potential energy, So

⇒ Potential energy = Kinetic energy

⇒ mgh = 60m

⇒ 10m × h = 60m [ g = 10 m/s² ]

⇒ 10h = 60

⇒ h = 6

Hence, The ball will rebound to a height of 6 m.

Answer 2

$\huge\bf{\underline{\underline{\gray{GIVEN:-}}}}$

• The ball is dropped from a height of 10m .

• After striking the ground the energy of the ball reduces by 40% .

$\huge\bf{\underline{\underline{\gray{TO\:FIND:-}}}}$

• The height that the ball can bounce back .

$\huge\bf{\underline{\underline{\gray{SOLUTION:-}}}}$

➪ Let,

• $\bf\red{Mass\:of\:the\:ball}$ = m

• $\bf\red{h}$ = 10m

• $\bf\red{g}$ = 10 m/s²

☞︎︎︎ We know that,

✞︎ ᴘᴏᴛᴇɴᴛɪᴀʟ ᴇɴᴇʀɢʏ ᴀᴛ ʜᴇɪɢʜᴛ $\bf\red{h}$ = $\mathcal\purple{m\:g\:h}$

⇝ᴘᴏᴛᴇɴᴛɪᴀʟ ᴇɴᴇʀɢʏ = m × 10 × 10

⇝ᴘᴏᴛᴇɴᴛɪᴀʟ ᴇɴᴇʀɢʏ = 100m J

☯︎ According to the question,

$\huge\red\checkmark$ On striking the ground level, the ball losses 40% of it's initial energy .

$\bf{\implies\:40\%\:of\:100m\:}$

$\rm{\implies\:\dfrac{40}{100}\times{100m}\:}$

$\sf\blue{\implies\:40m\:J\:}$

$\huge\red\checkmark$ The energy left on striking the ground is,

⇛ (100m - 40m) J

⇛ 60m J

➪ Let,

• The height of the ball after striking is $\bf\pink{h_1}$ .

✯ So final energy of the ball is,

$\mathcal\purple{\implies\:m\:g\:h_1\:=\:\bf{60m}\:}$

$\bf{\implies\:h_1\:=\:\dfrac{60m}{m\:g}\:}$

$\rm{\implies\:h_1\:=\:\dfrac{60m}{m\times{10}}\:}$

$\rm{\implies\:h_1\:=\:\dfrac{60m}{10m}\:}$

$\mathcal\green{\implies\:h_1\:=\:6m\:}$

$\huge\red\therefore$ The ball will bounce back to a height of '6m' .