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3 years ago

7

A ball is thrown into the air at an initial velocity of 18 meters per second from an initial height of 10 meters. The equation t

hat models the path of the ball is given by h=-4.9t^2+18t+10h How high is the ball at 3 seconds?
Group of answer choices

19.9 meters

108.1 meters

10 meters

The ball is no longer in the air at 3 seconds.

1 answer:

Sergio039 [100]3 years ago

7 0

Answer:

Step-by-step explanation:

This is your position equation:

$s(t)=-4.9t^2+18t+10$

There's a whole lot of information in that equation, but what we are concerned about right now is the height of the ball after t = 3 seconds. If this is the position of the ball at any time t, we will sub in 3 for t to find out where the ball is at 3 seconds.

$s(3)=-4.9(3^2)+18(3)+10$

which simplifies to

s(3) = -44.1 + 54 + 10 which is

s(3) = 19.9 meters

That's how high the ball is in the air at 3 seconds.

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Answer:

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Step-by-step explanation:

Given

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This is calculated as:

$P_{20} = 20 * \frac{N +1}{100}$

$P_{20} = 20 * \frac{8 +1}{100}$

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$P_{20} = \frac{20 * 9}{100}$

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This is then calculated as:

$P_{20} = 1st\ Item +0.8(2nd\ Item - 1st\ Item)$

$P_{20} = 16 + 0.8*(21 - 16)$

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Solving (b): The 25th percentile

This is calculated as:

$P_{25} = 25 * \frac{N +1}{100}$

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$P_{25} = \frac{25 * 9}{100}$

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$P_{75} = 28 + 0.75(2)$

$P_{75} = 28 + 1.5$

$P_{75} = 29.5$

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