Water flows from a tank of constant cross-sectional area 50

\mathrm{ft}^{2}

through an orifice of constant cross-sectional area

\frac{1}{4}

\mathrm{ft}^{2}

located at the bottom of the tank. Initially, the height of the water in the tank was 20 ft, and t sec later it was given by the equation

2 \sqrt{h}+\frac{1}{25} t-2 \sqrt{20}=0 \quad \quad \quad 0 \leq t \leq 50 \sqrt{20}

How fast was the height of the water decreasing when its height was 2 ft? $Water flows from a tank of constant cross-sectional area 50 \mathrm{ft}^{2} through an orifice of constant cross-sectional area \frac{1}{4} \mathrm{ft}^{2} located at the bottom of the tank. Initially, the height of the water in the tank was 20 ft, and t sec later it was given by the equation 2 \sqrt{h}+\frac{1}{25} t-2 \sqrt{20}=0 \quad \quad \quad 0 \leq t \leq 50 \sqrt{20} How fast was the height of the water decreasing when its height was 2 ft? A) 100 \sqrt{5}-50 \sqrt{2} ft/sec B) 100 \sqrt{5}-50 \sqrt{2} ft/sec. C) \frac{2}{25} ft/sec D) \frac{\sqrt{2}}{25} ft/sec$
A)

100 \sqrt{5}-50 \sqrt{2}

ft/sec
B)

100 \sqrt{5}-50 \sqrt{2}

ft/sec.
C)

\frac{2}{25}

ft/sec
D)

\frac{\sqrt{2}}{25}

ft/sec

The Equation of Motion Is Given for a Particle, Where 2.52.52.5

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The Equation of Motion Is Given for a Particle, Where $2.5$

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