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(A) Given, initial velocity $\overrightarrow{u} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} 	ext{ m/s}$.Height of the tower $h = 30 	ext{ m}$. Since $\hat{k

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$15$

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(A) Given, initial velocity $\overrightarrow{u} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} 	ext{ m/s}$.Height of the tower $h = 30 	ext{ m}$. Since $\hat{k}$ is vertically upward, the vertical displacement $S_z = -30 	ext{ m}$ and acceleration $a_z = -g = -10 	ext{ m/s}^2$.The initial vertical velocity component is $u_z = 5 	ext{ m/s}$.Using the equation of motion $S_z = u_z t + \frac{1}{2} a_z t^2$:$-30 = 5t - \frac{1}{2} 	imes 10 	imes t^2$$-30 = 5t - 5t^2$$t^2 - t - 6 = 0$$(t - 3)(t + 2) = 0$Since time cannot be negative, $t = 3 	ext{ s}$.In the horizontal plane, the velocity components are $v_x = 3 	ext{ m/s}$ (East) and $v_y = 4 	ext{ m/s}$ (North).The horizontal displacements in $3 	ext{ s}$ are:$x = v_x 	imes t = 3 	imes 3 = 9 	ext{ m}$$y = v_y 	imes t = 4 	imes 3 = 12 	ext{ m}$The horizontal range $R$ is the distance from the base of the tower:$R = \sqrt{x^2 + y^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 	ext{ m}$.

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