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omoplata

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## Homework Statement

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This is problem 10.27 in the Exercises for the Feynman Lectures on Physics, or Problem C-2 in Chapter 10 of Leighton and Vogt's Exercises in Introductory Physics.

## Homework Equations

Using ##v_f^2 = v_i^2 + 2 a S## for the water in the vertical direction, we can find the initial velocity ##v_i## of the water. ##v_f## is the final velocity, ##a## is the acceleration (or in this case the deceleration), ##S## is the distance traveled.

I can use ##P = x \rho g## to find the pressure difference the pump has to overcome, where ##x## is the height the water has to climb, ##\rho## is the density of water, and ##g## is the gravitational field.

##F = P A## to find the force necessary, where ##A## is the cross sectional area of the nozzle.

##Q = F v## to find the rate of work done on the water, where ##v## is the velocity of the water.

Then I can divide by 60% to find the actual power consumption of the motor.

## The Attempt at a Solution

The initial vertical velocity is ##v_i = v \sin(30^{\circ}) = v/2##. The vertical velocity at the apex is 0. Acceleration is ##-g##. Vertical distance is ##h##. Therefore,

$$0 = \left( \frac v 2 \right) ^2 - 2\ g\ h

\\v = \sqrt{8\ g\ h}$$

So I can find the velocity ##v## of the water. To find the pressure difference,

$$P = x\ \rho\ g$$

To find the force,

$$F = P\ A = x\ \rho\ g\ A$$

To find the power expended on the water,

$$Q = F\ v = x\ \rho\ g\ A\ \sqrt{8\ g\ h}$$

I'm going to use the following conversion factors.

$$1\ \rm{sq\ in} = 6.45 \times 10^{-4}\ \rm{m^2}

\\1\ \rm{ft} = 3.05 \times 10^{-1}\ \rm{m}

\\1\ \rm{lb\ ft^{-3}} = 1.60 \times 10^1 \rm{kg\ m^{-3}}$$

After calculation, I get ##Q = 1.66 \times 10^3 \rm{W}##. After dividing by 60%, I get ##2.77 \times 10^3 \rm{W}##, or ##2.77\rm{kW}##. The answer given in the book is ##25\rm{kW}##. Where did I go wrong?