Answer
The maximum height reached by the 2.00-kg object is 1.457 meters.
Work Step by Step
Let $m_1 = 5.00~kg$ and let $m_2 = 2.00~kg$
We can find the acceleration of the system.
$\sum F = (m_1 + m_2)~a$
$m_1~g - m_2~g = (m_1 + m_2)~a$
$a = \frac{m_1~g - m_2~g}{m_1 + m_2}$
$a = \frac{(5.00~kg)(9.80~m/s^2) - (2.00~kg)(9.80~m/s^2)}{5.00~kg + 2.00~kg}$
$a = 4.20~m/s^2$
After the system starts moving, the 2.00-kg object will accelerate upward for 0.600 meters and then the 5.00-kg object will hit the ground. We can find the speed of the 2.00-kg object after it accelerates for 0.600 meters.
$v^2 = v_0^2+2ay = 0 + 2ay$
$v = \sqrt{2ay}= \sqrt{(2)(4.20~m/s^2)(0.600~m)}$
$v = \sqrt{5.04~m^2/s^2}$
At this point, the 2.00-kg object will continue moving upward until gravity brings it to rest at its maximum height. We can find the additional distance $y$ that the object goes up. We can let $v_0 = \sqrt{5.04~m^2/s^2}$ for this part of the question.
$y = \frac{v^2-v_0^2}{2a} = \frac{0-(5.04~m^2/s^2)}{(2)(-9.80~m/s^2)}$
$y = 0.257~m$
We can find the maximum height reached by the 2.00-kg object.
$0.600~m + 0.600~m + 0.257~m = 1.457~m$
The maximum height reached by the 2.00-kg object is 1.457 meters.
