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3 years ago

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An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. how high does it rise (v = 0 cm

/s at top of rise). remember, acceleration is -9.80 m/s2.

2 answers:

maw [93]3 years ago

7 0

Answer: The maximum height is 20.4 meters

Explanation: We have that the initial velocity is 20m/s and the initial position is 0m.

The acceleration of the object is:

a(t) = -9.8m/s^2

for the velocity, we integrte over time.

v(t) = -9.8m/s^2*t + 20m/s

and the position is:

p(t) = -4.9m/s^2*t^2 + 20m/s*t + 0

And we want to do the maximum height (in this case the maximum value of p(t))

First, we need to find the time in wich te velocity is equal to zero, which means that the object stops to going up and starts moving down.

v(t) = 0 = -9.8m/s^2*t + 20m/s

t = (20/9.8)s = 2.04s

Now, we put this time in the position equation:

p(2.04s) = -4.9m/s^2*(2.04s)^2 + 20m/s*2.04s = 20.4m

Aliun [14]3 years ago

5 0

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The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

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Using formula of force

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Where, W = water weight

V = volume

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$F=62.5\times(441\pi\times\Delta h)$

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We need to calculate the work done

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs$

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh$

$W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(10\times10-\dfrac{100}{2}-0)$

$W=4329507.37572$

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh$

$W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(12\times10-\dfrac{100}{2}-0)$

$W=1929375\pi$

$W=6061310.32\ foot- pound$

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

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