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 $12.4m$ to have it bounce back to the same height (ignore resistance)?

• A. $6.55m/s$
• B. $5.66m/s$
• C. $2.50m/s$
• D. $7.50m/s$

Answer 1

The correct option is A $6.55m/s$

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
$v=0,u=v_2$
$\therefore v^2=u^2-2gh$
$0=\frac{85}{100}(u^2+243.04)-2\times 9.8\times 12.4$
$\frac{85}{100}{u}^2=36.46$
$u^2=\frac{36.46\times 100}{85}=42.89$
$u=6.55m/s$