Question

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)?

• A. 6.55 m/s
• B. 12.0 m/s
• C. 8.6 m/s
• D. 4.55 m/s

Answer 1

The correct option is A 6.55 m/s

Given $h=12.4,v=?$
$\therefore v^2=u^2+2gh$
$i.e.,v^2=u^2+2\times 9.8\times 12.4$
$=u^2+243.04$
Kinetic energy of the ball when it just hits the wall
$=\frac{1}{2}{mv}^2=\frac{1}{2}m(u^2+243.04)$
The K.E. of ball after the impact
$=\frac{(100-15)}{100}\times \frac{1}{2}m(u^2+243.04)$
$=\frac{85}{100}\times \frac{1}{2}m(u^2+243.04)$
Let $v^2$ be the upward velocity just after the collision with the ground.
So, $\frac{1}{2}m{v}_2^2=\frac{85}{100}\times \frac{1}{2}m(u^2+243.04)$
$v_2^2=\frac{85}{100}(u^2+243.04)$
Now, taking upward motion
$v=0,u=v_2$
$\therefore v^2=u^2-2gh$
$0=\frac{85}{100}(u^2+243.04)-2\times 9.8\times 12.4$
$\frac{85}{100}{u}^2=36.46$
$u^2=\frac{36.46\times 100}{85}=42.89$
$u=6.55m/s$