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(D) The time taken for the first fall is $t_0 = \sqrt{\frac{2h}{g}}$.After the first impact, the ball rises to height $h_1 = e^2 h$ and falls back, ta

Vedclass · Accepted Answer

$\frac{17}{18}$

Vedclass · Answer

(D) The time taken for the first fall is $t_0 = \sqrt{\frac{2h}{g}}$.After the first impact, the ball rises to height $h_1 = e^2 h$ and falls back, taking time $t_1 = 2 \sqrt{\frac{2h_1}{g}} = 2e \sqrt{\frac{2h}{g}}$.Similarly, for subsequent bounces, the time taken is $t_n = 2e^n \sqrt{\frac{2h}{g}}$.The total time $T$ is the sum of the initial fall and all subsequent bounces:$T = \sqrt{\frac{2h}{g}} + 2e \sqrt{\frac{2h}{g}} + 2e^2 \sqrt{\frac{2h}{g}} + \dots = \sqrt{\frac{2h}{g}} \left( 1 + 2e + 2e^2 + \dots ight)$.Using the sum of a geometric series for $e Given $h = 0.4 	ext{ m}$, $g = 10 	ext{ m/s}^2$, and $T = 10 	ext{ s}$:$10 = \sqrt{\frac{2 	imes 0.4}{10}} \left( \frac{1+e}{1-e} ight) = \sqrt{0.08} \left( \frac{1+e}{1-e} ight) = 0.2828 \left( \frac{1+e}{1-e} ight)$.Actually, using $g = 9.8 	ext{ m/s}^2$ or simplifying the expression: $10 = \sqrt{0.08} \frac{1+e}{1-e} \implies \frac{10}{0.2828} = \frac{1+e}{1-e} \approx 35.35$.Solving for $e$ yields $e \approx 0.944$, which corresponds to $\frac{17}{18} \approx 0.944$.

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