A ball falls freely from rest on to a hard ho | vedclass

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(C) Given: Initial velocity $u = 7 	ext{ m/s}$, coefficient of restitution $e = 0.75$, acceleration due to gravity $g = 10 	ext{ m/s}^2$.Initial hei

Vedclass · Accepted Answer

$8.75$

Vedclass · Answer

(C) Given: Initial velocity $u = 7 	ext{ m/s}$, coefficient of restitution $e = 0.75$, acceleration due to gravity $g = 10 	ext{ m/s}^2$.Initial height $H = \frac{u^2}{2g} = \frac{7^2}{2 	imes 10} = \frac{49}{20} = 2.45 	ext{ m}$.The ball bounces back with velocity $v_1 = eu = 0.75 	imes 7 = 5.25 	ext{ m/s}$.The height reached after the first bounce is $H_1 = \frac{v_1^2}{2g} = e^2 H = (0.75)^2 	imes 2.45 = 0.5625 	imes 2.45 = 1.378125 	ext{ m}$.The total distance $S$ traveled by the ball is $S = H + 2H_1 + 2H_2 + \dots = H + 2(e^2H + e^4H + \dots) = H(1 + 2e^2(1 + e^2 + e^4 + \dots))$.Using the sum of an infinite geometric series $S_{\infty} = \frac{a}{1-r}$, we get $S = H(1 + 2e^2(\frac{1}{1-e^2})) = H(\frac{1-e^2+2e^2}{1-e^2}) = H(\frac{1+e^2}{1-e^2})$.Substituting the values: $S = 2.45 	imes \frac{1 + (0.75)^2}{1 - (0.75)^2} = 2.45 	imes \frac{1 + 0.5625}{1 - 0.5625} = 2.45 	imes \frac{1.5625}{0.4375} = 2.45 	imes 3.5714 = 8.75 	ext{ m}$.

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