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(C) Let the top of the tower be the origin $(0, 25)$ in the $xy$-plane. The masses are $m_1 = 12 \,kg$ and $m_2 = 6 \,kg$.The initial velocities are $

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$30$

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(C) Let the top of the tower be the origin $(0, 25)$ in the $xy$-plane. The masses are $m_1 = 12 \,kg$ and $m_2 = 6 \,kg$.The initial velocities are $\vec{v}_1 = 15 \hat{j} \,ms^{-1}$ and $\vec{v}_2 = 20 \hat{i} \,ms^{-1}$.The vertical component of the velocity of the centre of mass is $v_{cm,y} = \frac{m_1 v_{1y} + m_2 v_{2y}}{m_1 + m_2} = \frac{12(15) + 6(0)}{12 + 6} = \frac{180}{18} = 10 \,ms^{-1}$.The vertical acceleration of the centre of mass is $a_{cm,y} = -g = -10 \,ms^{-2}$.The height of the centre of mass from the top of the tower is $y_{cm} = v_{cm,y} t - \frac{1}{2} g t^2$.The maximum height from the top is reached when $v_{cm,y}(t) = 0$,i.e.,$10 - 10t = 0 \implies t = 1 \,s$.The maximum displacement from the top is $y_{max} = 10(1) - \frac{1}{2}(10)(1)^2 = 10 - 5 = 5 \,m$.The total height from the ground is $H = 25 + 5 = 30 \,m$.

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