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Question-213664




Question Number 213664 by 281981 last updated on 13/Nov/24
Answered by mr W last updated on 13/Nov/24
Commented by mr W last updated on 13/Nov/24
(x/(20−y))=(5/y)  x=((100)/y)−5  u=(dx/dt)=−((100)/y^2 )×(dy/dt)  (dy/dt)=v=(√(2gy))  ⇒u=−((100(√(2gy)))/y^2 )           =−((100(√(2×10×16)))/(16^2 ))≈−7 m/s
$$\frac{{x}}{\mathrm{20}−{y}}=\frac{\mathrm{5}}{{y}} \\ $$$${x}=\frac{\mathrm{100}}{{y}}−\mathrm{5} \\ $$$${u}=\frac{{dx}}{{dt}}=−\frac{\mathrm{100}}{{y}^{\mathrm{2}} }×\frac{{dy}}{{dt}} \\ $$$$\frac{{dy}}{{dt}}={v}=\sqrt{\mathrm{2}{gy}} \\ $$$$\Rightarrow{u}=−\frac{\mathrm{100}\sqrt{\mathrm{2}{gy}}}{{y}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{100}\sqrt{\mathrm{2}×\mathrm{10}×\mathrm{16}}}{\mathrm{16}^{\mathrm{2}} }\approx−\mathrm{7}\:{m}/{s} \\ $$

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