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Lesechka [4]
3 years ago
10

Brian "bicycle kicked" a soccer ball up from a height of 4ft off the ground with an initial upward velocity of 34 feet/sec. As t

he ball was coming down, Mike jumped up and at 8 ft off the ground, headed it into the goal and win the game. How long did it take for the ball to come down and reach Mike's head?
Mathematics
1 answer:
N76 [4]3 years ago
3 0

Answer:

1.95 seconds or 2.06 seconds

(Depending on your acceleration value)

Step-by-step explanation:

s = ut + 0.5at²

u = 34 ft/s

a = -10 m/s² = -32.8084 ft/s²

s = 8-4 = 4 ft

4 = 34t + ½(-32.8084)t²

4 = 34t - 16.4042t²

16.4042t² - 34t + 4 = 0

t = 1.946, 0.125

Mike catches the ball the second time it was at a height of 8ft.

Do t = 1.95 s

If you use a = 32.17405 ft/s²,

t = 2.05655765 (2.06 seconds in 3 sf)

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Answer:

\bar x = 260.1615

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Step-by-step explanation:

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n =20

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

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\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}

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This is calculated as:

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Solving (c): 95% confidence interval of standard deviation

We have:

c =0.95

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df = 19

Determine the critical value at row df = 19 and columns \frac{\alpha}{2} and 1 -\frac{\alpha}{2}

So, we have:

X^2_{0.025} = 32.852 ---- at \frac{\alpha}{2}

X^2_{0.975} = 8.907 --- at 1 -\frac{\alpha}{2}

So, the confidence interval of the standard deviation is:

\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} } to \sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }

70.69 * \sqrt{\frac{20 - 1}{32.852} to 70.69 * \sqrt{\frac{20 - 1}{8.907}

70.69 * \sqrt{\frac{19}{32.852} to 70.69 * \sqrt{\frac{19}{8.907}

53.76 to 103.25

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