A ball is dropped from a height of $10\ m$. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? $(g=10\ m s^{-2})$

As given, Height $h_1=10\ m$

$g=10\ ms^{-2}$

Let $m$ be the mass of the ball.

Therefore, potential energy $P.E.=mgh_1=m\times 10\times 10=100\times m\ J$

After reaching the ground, it losses its $40$% energy, so the remaining energy is $60$% of P.E.

The remained energy $=60$% of P.E.

$=\frac{60}{100}\times100\times m$

$=60\times m\ J$

Let the ball bounce at $h_2$ height with the remaining energy.

Therefore, $m\times g\times h_2=60\times m$

Or $h_2=\frac{60\times m}{g\times m}$

Or $h_2=\frac{60\times m}{10\times m}$

Or $h_2=6\ meter$

Therefore, the ball will bounce $6\ meter$ with the remaining energy.


Simply Easy Learning

Updated on: 10-Oct-2022


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