A juggler maintains four balls in vertically | vedclass

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(C) The juggler maintains $4$ balls in the air. The time interval between throwing consecutive balls is $T = 1/4 \, s$.Since there are $4$ balls,the t

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$1.25$

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(C) The juggler maintains $4$ balls in the air. The time interval between throwing consecutive balls is $T = 1/4 \, s$.Since there are $4$ balls,the total time a ball stays in the air (time of flight) must be $T_{flight} = 4 	imes T = 4 	imes (1/4) = 1 \, s$.The time taken to reach the maximum height is $t = T_{flight} / 2 = 1 / 2 = 0.5 \, s$.Using the equation of motion for a ball thrown vertically upward,at maximum height,the final velocity $v = 0$.Using $v = u - gt$,we get $0 = u - (10)(0.5)$,so $u = 5 \, m/s$.The maximum height $H$ reached is given by $H = \frac{u^2}{2g} = \frac{5^2}{2 	imes 10} = \frac{25}{20} = 1.25 \, m$.Alternatively,using $H = \frac{1}{2}gt^2$ for the downward journey from the peak: $H = \frac{1}{2} 	imes 10 	imes (0.5)^2 = 5 	imes 0.25 = 1.25 \, m$.

$A$ juggler maintains four balls in vertically upward motion. He throws the next ball every $1/4 \, s$. For the show to continue,what should be the height to which he throws each ball? (Assume $g = 10 \, m/s^2$)

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