Consider the configuration of a stationary wa | vedclass

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(C) The velocity of efflux from the hole is given by $v_{e} = \sqrt{2gh}$.The horizontal range $x$ of the water jet is given by $x = v_{e} 	imes t$,w

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(C) The velocity of efflux from the hole is given by $v_{e} = \sqrt{2gh}$.The horizontal range $x$ of the water jet is given by $x = v_{e} 	imes t$,where $t$ is the time of fall from height $H$.Since $H = \frac{1}{2}gt^{2}$,we have $t = \sqrt{\frac{2H}{g}}$.Thus,$x = \sqrt{2gh} 	imes \sqrt{\frac{2H}{g}} = 2\sqrt{hH}$.The speed $v$ of the bucket must match the rate of change of the range $x$ to keep the water falling into it:$v = \frac{dx}{dt} = \frac{d}{dt}(2\sqrt{hH}) = 2\sqrt{H} \frac{d}{dt}(\sqrt{h}) = 2\sqrt{H} \cdot \frac{1}{2\sqrt{h}} \cdot \frac{dh}{dt} = \sqrt{\frac{H}{h}} \cdot \frac{dh}{dt}$.From the equation of continuity,$A_{0} \left(-\frac{dh}{dt}ight) = A v_{e} = A \sqrt{2gh}$.Therefore,$\frac{dh}{dt} = -\frac{A}{A_{0}} \sqrt{2gh}$.Substituting this into the expression for $v$ (taking magnitude): $v = \sqrt{\frac{H}{h}} \cdot \frac{A}{A_{0}} \sqrt{2gh} = \frac{A}{A_{0}} \sqrt{2gH}$.Given $A = 5 \, cm^{2}$,$A_{0} = 500 \, cm^{2}$,$g = 10 \, m/s^{2}$,and $H = 5 \, m$:$v = \left(\frac{5}{500}ight) \sqrt{2 	imes 10 	imes 5} = \frac{1}{100} 	imes \sqrt{100} = \frac{10}{100} = 0.1 \, m/s$.

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