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(B) Let $h$ be the height of the tower and $u$ be the initial velocity.For ball $A$ thrown vertically upwards: $h = -ut_A + \frac{1}{2}gt_A^2$ (taking

Vedclass · Accepted Answer

$3.46$

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(B) Let $h$ be the height of the tower and $u$ be the initial velocity.For ball $A$ thrown vertically upwards: $h = -ut_A + \frac{1}{2}gt_A^2$ (taking downward direction as positive,$h = -ut_A + \frac{1}{2}gt_A^2$ is incorrect,let's use standard kinematic equation $h = -ut + \frac{1}{2}gt^2$ where $h$ is displacement). Correct approach: Let $h$ be the height of the tower. For ball $A$ (upwards): $h = -ut_A + \frac{1}{2}gt_A^2 \implies u = \frac{1}{2}gt_A - \frac{h}{t_A}$.For ball $B$ (downwards): $h = ut_B + \frac{1}{2}gt_B^2 \implies u = \frac{h}{t_B} - \frac{1}{2}gt_B$.Equating $u$: $\frac{1}{2}gt_A - \frac{h}{t_A} = \frac{h}{t_B} - \frac{1}{2}gt_B$.$\frac{1}{2}g(t_A + t_B) = h(\frac{1}{t_A} + \frac{1}{t_B}) = h(\frac{t_A + t_B}{t_A t_B})$.Thus,$h = \frac{1}{2}gt_A t_B$.For ball $C$ (horizontally): The vertical displacement is $h = \frac{1}{2}gt_C^2$.Comparing the two expressions for $h$: $\frac{1}{2}gt_C^2 = \frac{1}{2}gt_A t_B$.$t_C^2 = t_A t_B \implies t_C = \sqrt{t_A t_B}$.Given $t_A = 6 \ s$ and $t_B = 2 \ s$:$t_C = \sqrt{6 	imes 2} = \sqrt{12} \approx 3.46 \ s$.

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