A ball loses $15.0\%$ of its kinetic energy when it bounces back from a concrete wall. With what speed you must throw it vertically down from a height of $12.4\, m$ to have it bounce back to the same height (ignore air resistance)? ............. $\mathrm{m} / \mathrm{s}$
A ball loses $15.0\%$ of its kinetic energy when it bounces back from a concrete wall. With what speed you must throw it vertically down from a height of $12.4\, m$ to have it bounce back to the same height (ignore air resistance)? ............. $\mathrm{m} / \mathrm{s}$
- [AIIMS 2010]
$\begin{gathered}
Given:\,h = 12.4,v = ? \hfill \\
\therefore \,{v^2} = {u^2} + 2gh \hfill \\
i.e.,\,{v^2} = {u^2} + 2 \times 9.8 \times 12.4 \hfill \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {u^2} + 243.04 \hfill \\
\end{gathered} $
Kinetic energy of the ball when it just hits
the wall
$ = \frac{1}{2}m{v^2} = \frac{1}{2}m\left( {{u^2} + 243.04} \right)$
The $K.E.$ of ball after the impact
$ = \frac{{\left( {100 - 15} \right)}}{{100}} \times \frac{1}{2}m\left( {{u^2} + 243.04} \right)$
$ = \frac{{85}}{{100}} \times \frac{1}{2}m\left( {{u^2} + 243.04} \right)$
Let $v_2$ be the upward velocity just after the collision with the ground.
So, $\frac{1}{2}mv_2^2 = \frac{{85}}{{100}} \times \frac{1}{2}m\left( {{u^2} + 243.04} \right)$
$v_2^2 = \frac{{85}}{{100}}\left( {{u^2} + 243.04} \right)$
Now, taking upward motion
$ v = 0, u = v_2$
$\therefore \,{v^2} = {u^2} - 2gh$
$0 = \frac{{85}}{{100}}\left( {{u^2} + 243.04} \right) - 2 \times 9.8 \times 12.4$
$\begin{array}{l}
\,{u^2} = \frac{{36.46 \times 100}}{{85}} = 42.89\\
\,\,\,\,\,\,\,\,\,u = 6.55\,m/s
\end{array}$
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