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Question Number 68510 by oyemi kemewari last updated on 12/Sep/19

	Answered by mr W last updated on 14/Sep/19

	$v = \sqrt{2 g h}$ $b a s e a r e a o f t a n k s = A = 1 {f t}^{2}$ $A \frac{d h}{2} = - \frac{π d^{2}}{4} v d t$ $d h = - \frac{π d^{2} \sqrt{2 g h}}{2 A} d t$ $\frac{d h}{\sqrt{h}} = - \frac{π d^{2} \sqrt{2 g}}{2 A} d t$ $\int_{h_{0}}^{h} \frac{d h}{\sqrt{h}} = - \frac{π d^{2} \sqrt{2 g}}{2 A} \int_{0}^{t} d t$ $2 [\sqrt{h} - \sqrt{h_{0}}] = - \frac{π d^{2} \sqrt{2 g}}{2 A} t$ $\sqrt{h_{0}} - \sqrt{h} = \frac{π d^{2} \sqrt{2 g}}{4 A} t$ $\Rightarrow t = \frac{4 A (\sqrt{h_{0}} - \sqrt{h})}{π d^{2} \sqrt{2 g}}$ $h_{0} = 1 f t$ $d = 0.5 i n = \frac{1}{24} f t$ $f o r h = 0.25 f t :$ $t = \frac{4 \times 1 (\sqrt{1} - \sqrt{0.25})}{π {(\frac{1}{24})}^{2} \sqrt{2 g}} = 82 s e c o n d s$ $f o r h = 0 :$ $t = \frac{4 \times 1 (\sqrt{1} - \sqrt{0})}{π {(\frac{1}{24})}^{2} \sqrt{2 g}} = 164 s e c o n d s$
	Commented by mr W last updated on 14/Sep/19

	Commented by oyemi kemewari last updated on 17/Sep/19


	Commented by oyemi kemewari last updated on 28/Sep/19
	thank you sir
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