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(D) The tower height is $H = 180 \,m$. The balls meet at $100 \,m$ above the ground,so they have traveled a distance $S = 180 - 100 = 80 \,m$ from the

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$30$

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(D) The tower height is $H = 180 \,m$. The balls meet at $100 \,m$ above the ground,so they have traveled a distance $S = 180 - 100 = 80 \,m$ from the top.For ball $A$,released from rest $(u_A = 0)$:$S = \frac{1}{2} g t_A^2 \implies 80 = \frac{1}{2} 	imes 10 	imes t_A^2 \implies t_A^2 = 16 \implies t_A = 4 \,s$.Ball $B$ is thrown at $t = 2 \,s$,so its travel time is $t_B = t_A - 2 = 4 - 2 = 2 \,s$.For ball $B$,using the equation of motion $S = u t_B + \frac{1}{2} g t_B^2$ (taking downward as positive):$80 = u(2) + \frac{1}{2}(10)(2)^2$$80 = 2u + 20$$2u = 60$$u = 30 \,m \,s^{-1}$.

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