Physics, asked by STOKESY, 3 months ago

A ball of mass 10 gram is dropped from a height of hundred metre, after striking the ground 25% of kinetic energy decreases. how much height the ball will attend after striking the ground?​

Answers

Answered by shadowsabers03
10

By energy conservation, the kinetic energy of the ball when it hits the ground is equal to its potential energy at the height where the ball is dropped.

[The ball has no kinetic energy at the height where it is dropped, and no potential energy at ground.]

That is,

\sf{\longrightarrow KE=mgh}

Here,

  • \sf{m=0.01\ kg}
  • \sf{h=100\ m}

Let \sf{g=10\ m\,s^{-2}.} Then,

\sf{\longrightarrow KE=0.01\times10\times100}

\sf{\longrightarrow KE=10\ J}

After striking the ground the kinetic energy gets reduced by 25%. So the new kinetic energy becomes,

\sf{\longrightarrow KE=10\times\dfrac{75}{100}}

\sf{\longrightarrow KE=7.5\ J}

This is equal to the potential energy at the height till which the ball rebounds, i.e.,

\sf{\longrightarrow mgh'=7.5}

\sf{\longrightarrow 0.01\times10h'=7.5}

\sf{\longrightarrow 0.1h'=7.5}

\sf{\longrightarrow\underline{\underline{h'=75\ m}}}

Hence the ball will rise by 75 m after striking the ground.

Answered by Anonymous
80

Given,

  • M = 10g = 0.01kg
  • h = 100m
  • After striking the ground 25% of kinetic energy decreases.

To Find,

  • How much height the ball will attend after striking the ground?

Solution,

:\implies Kinetic  \:  \: Energy = Mgh \:  \: (g = 10m {s}^{ - 2} ) \\  \\ :\implies Kinetic  \:  \: Energy = 0.01kg \times 10m {s}^{ - 2}  \times 100m \\  \\ :\implies Kinetic  \:  \: Energy = 10J

After striking the ground 25% of kinetic energy decreases.

So,

75% of Kinetic Energy remains

:\implies 75 \: percent \:  \: of \:  \: 10J \\  \\  :\implies \frac{75}{100}  \times 10J \\  \\:\implies  7.5J

Required Answer,

:\implies Kinetic  \:  \: Energy = 7.5J \\  \\:\implies  Mgh = 7.5J \\  \\:\implies  0.01kg \times 10m {s}^{ - 2}  \times h = 7.5J \\  \\:\implies  0.1kgm {s}^{ - 2}  \times h = 7.5J \\  \\ :\implies h =  \frac{7.5J}{0.1kgm {s}^{ - 2} }  \\  \\ :\implies h = 75m

Height = 75m

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