Answer
The smaller ball reaches a height of $~~7.2~m$
Work Step by Step
We can use conservation of energy to find the speed after falling from a height of $1.8~m$:
$\frac{1}{2}m_1v^2 = m_1~gh$
$v^2 = 2gh$
$v = \sqrt{2gh}$
$v = \sqrt{(2)(9.8~m/s^2)(1.8~m)}$
$v = 5.94~m/s$
At the moment of the collision, we can let the velocity of the larger ball be $v_{1i} = 5.94~m/s$ and the velocity of the smaller ball be $v_{2i} = -5.94~m/s$
In part (a), we found that $m = 0.21~kg$
We can use Equation (9-76) to find the velocity $v_{2f}$ of the smaller ball immediately after the collision:
$v_{2f} = \frac{2M}{M+m}~v_{1i}+\frac{m-M}{M+m}~v_{2i}$
$v_{2f} = \left[\frac{(2)(0.63~kg)}{0.63~kg+0.21~kg}\right]~(5.94~m/s)+\left[\frac{0.21~kg-0.63~kg}{0.63~kg+0.21~kg}\right]~(-5.94~m/s)$
$v_{2f} = (1.5)~(5.94~m/s)+(-0.5)~(-5.94~m/s)$
$v_{2f} = 11.88~m/s$
In the next part of the solution, we can let $v_i = 11.88~m/s$ and $v_f = 0$
We can find the height reached by the smaller ball:
$v_f^2 = v_i^2+2ay$
$y = \frac{v_f^2 - v_i^2}{2a}$
$y = \frac{0 - (11.88~m/s)^2}{(2)(-9.8~m/s^2)}$
$y = 7.2~m$
The smaller ball reaches a height of $~~7.2~m$
