A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by ℎ = −16푡 ଶ+103푡+5 where t is time (in seconds).

Answers

Answer 1

Question:

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by h(t)=-16t^2+80t+5, where t is measured in seconds.  Write an equation to determine how long it will take for the ball to reach the ground.

Answer:

[tex]t = 5.0625[/tex]

Step-by-step explanation:

Given

[tex]h(t)=-16t^2+80t+5[/tex]

Required

Find t when the ball hits the ground

This implies that h(t) = 0

So, we have:

[tex]0=-16t^2+80t+5[/tex]

Reorder

[tex]-16t^2+80t+5 = 0[/tex]

Using quadratic formula, we have:

[tex]t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

Where

[tex]a = -16[/tex]      [tex]b =80[/tex]      [tex]c = 5[/tex]

So, we have:

[tex]t = \frac{-80\±\sqrt{80^2 - 4*-16*5}}{2*-16}[/tex]

[tex]t = \frac{-80\±\sqrt{6400 +320}}{-32}[/tex]

[tex]t = \frac{-80\±\sqrt{6720}}{-32}[/tex]

[tex]t = \frac{-80\±82.0}{-32}[/tex]

This gives:

[tex]t = \frac{-80+82.0}{-32}[/tex] or [tex]t = \frac{-80-82.0}{-32}[/tex]

[tex]t = \frac{2}{-32}[/tex] or [tex]t = \frac{-162}{-32}[/tex]

[tex]t = -\frac{2}{32}[/tex] or [tex]t = \frac{162}{32}[/tex]

But time can not be negative.

So, we have:

[tex]t = \frac{162}{32}[/tex]

[tex]t = 5.0625[/tex]

Hence, time to hit the ground is 5.0625 seconds


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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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OR

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Answers

Answer:

[tex]C(w) = 5w + 12[/tex]

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Step-by-step explanation:

Given information:

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- - - -

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Answers

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Answers

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