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Question

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

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Solution

## Step 1: Given DataInitial height from where the ball is dropped $h=10m$Acceleration due to gravity $g=10m{s}^{-2}$The energy of the ball reduced $=40%$Let $m$ be the mass of the ball.Let the height of the ball after rebound be $h\text{'}$.Step 2: Calculate the total potential energyWe know that the initial potential energy of the ball is$PE=mgh$ $=m×10×10$ $=100m$Step 3: Calculate the reduced potential energySince $100m$ is the total energy, then the reduced energy can be given as$=100m-\left(40%of100m\right)\phantom{\rule{0ex}{0ex}}=100m-\left(\frac{40}{100}×100m\right)\phantom{\rule{0ex}{0ex}}=100m-40m\phantom{\rule{0ex}{0ex}}=60m$Step 4: Calculate the bounced heightThis is the energy of the ball when it bounces back.$\therefore 60m=mgh\text{'}$$⇒h\text{'}=\frac{60}{g}$$⇒h\text{'}=\frac{60}{10}=6m$Hence, the ball can bounce back $6m$ high.

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