Answer
Gerald threw the ball of putty with a speed of $~30.4~m/s$
Work Step by Step
Let $m_p$ be the mass of the putty. Let $m_t$ be the mass of the target.
We can use conservation of energy to find the speed $v_f$ of the target and the putty just after the collision:
$KE = U_g$
$\frac{1}{2}(m_t+m_p)~v_f^2 = (m_t+m_p)~gh$
$v_f = \sqrt{2gh}$
$v_f = \sqrt{(2)(9.80~m/s^2)(1.50~m)}$
$v_f = 5.422~m/s$
We can use conservation of momentum to find the speed $v_0$ of the putty before the collision:
$m_p~v_0 = (m_t+m_p)~v_f$
$v_0 = \frac{(m_t+m_p)~v_f}{m_p}$
$v_0 = \frac{(2.30~kg+0.50~kg)~(5.422~m/s)}{0.50~kg}$
$v_0 = 30.4~m/s$
Gerald threw the ball of putty with a speed of $~30.4~m/s$.
