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Nostrana [21]
3 years ago
15

A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

t+4 . Suppose the juggler missed and ball hit the ground . Find the maximum height of the ball and time it took to reach the ground. Round all answers to the nearest hundredth .
Mathematics
1 answer:
malfutka [58]3 years ago
8 0
PART A

The given equation is

h(t) = - 16 {t}^{2} + 40t + 4

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4

We add and subtract

- 16(- \frac{5}{4} )^{2}

to get,

h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4

We again factor -16 out of the first two terms to get,

h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4

This implies that,

h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4

The quadratic trinomial above is a perfect square.

h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4

This finally simplifies to,

h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29

The vertex of this function is

V( \frac{5}{4} ,29)

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

29

PART B

When the ball hits the ground,

h(t) = 0

This implies that,

- 16 ( t- \frac{5}{4}) ^{2} +29 = 0

We add -29 to both sides to get,

- 16 ( t- \frac{5}{4}) ^{2} = - 29

This implies that,

( t- \frac{5}{4}) ^{2} = \frac{29}{16}

t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }

t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}

t = \frac{ 5 + \sqrt{29} }{4} = 2.60

or

t = \frac{ 5 - \sqrt{29} }{4} = - 0.10

Since time cannot be negative, we discard the negative value and pick,

t = 2.60s
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Since you did not attach any picture we cannot say for sure what is the correct answer, but we can discuss the options in order to find the most probable correct answer.

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I guess the oblique cylinder has height h and it is divided into many (probably 10) cross-sections.

Option A: <span>πr2h
This is exactly the volume of the right cylinder, therefore, unless you are given a cross-section of height h (which would be too easy), this won't be the correct answer.

Option B: </span><span>4πr2h
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Option C: <span>1 10 πr2h
Here comes a n issue with the notation: I think the right number you meant to write is (1/10)</span>·πr2h and not 110·<span>πr2h.
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Option D: </span><span>1 2 πr2h
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