A fire hydrant delivers water of density $\rho$ at a volume rate $L$. The water travels vertically upward through the hydrant and then does $90^o $ turn to emerge horizontally at speed $V$. The pipe and nozzle have uniform crosssection throughout. The force exerted by the water on the corner of the hydrant is
Diffcult
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Change in momentum in vertical direction in time $\mathrm{t}=$ momentum final$-$momentum
initial in vertical direction $=-\rho L V t$
Change in momentum in horizontal direction in time $\mathrm{t}=$ momentum final$-$momentum
initial in horizontal direction $=\rho L V t$
So net change in momentum in time $\mathrm{t}=\sqrt{(\rho L V t)^{2}+(\rho L V t)^{2}}=\sqrt{2} \rho L V t$
Force $=$ Change in momentum per time $=\sqrt{2} \rho L V$
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