Mathematics, 19.01.2021 09:20 kyronthomas52

I would appreciate it if you could answer the questions please- A ball is thrown upward at 48ft/s from a building that is 100 feet high the function h(t)= -16+84t + 100 expresses the balls height, h in feet given the time t, in seconds since the ball was thrown. please answer the questions if you know the answer I would appreciate it

I would appreciate it if you could answer the questions please- A ball is thrown upward at 48ft/s f

Answers: 3

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Answers

Answer from: Nolife012603

Problem 8

1.5 seconds

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Work Shown:

h(t) = -16t^2+48t+100 is the same as y = -16x^2+48x+100, just with different variables.

Compare this to y = ax^2+bx+c to find that a = -16, b = 48, c = 100

Then compute the x coordinate of the vertex -b/(2a) getting

h = -b/(2a)

h = -48/(2*(-16))

h = 1.5

It takes 1.5 seconds for the ball to reach its peak height. The vertex represents the max height since the parabola opens downward.

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Problem 9

136 feet

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Work Shown:

Plug the result from problem 8 into the function

y = -16x^2 + 48x + 100

y = -16(1.5)^2 + 48(1.5) + 100

y = 136

The vertex is (1.5, 136)

This means at 1.5 seconds, the ball is 136 feet off the ground at its highest point.

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Problem 10

approximately 4.415 seconds

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Work Shown:

Plug in y = 0 and solve for x

y = -16x^2 + 48x + 100

0 = -16x^2 + 48x + 100

-16x^2 + 48x + 100 = 0

Apply the quadratic formula

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(48)\pm\sqrt{(48)^2-4(-16)(100)}}{2(-16)}\\\\x = \frac{-48\pm\sqrt{8704}}{-32}\\\\x \approx \frac{-48\pm93.29523032}{-32}\\\\x \approx \frac{-48+93.29523032}{-32}\ \text{ or } \ x \approx \frac{-48-93.29523032}{-32}\\\\x \approx \frac{45.29523032}{-32}\ \text{ or } \ x \approx \frac{-141.2952303}{-32}\\\\x \approx -1.41547593\ \text{ or } \ x \approx 4.41547595\\\\$

A negative x value makes no sense because time cannot be negative, so we ignore the first solution. The only practical solution is roughly x = 4.415

It takes approximately 4.415 seconds for the ball to reach the ground.

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Problem 11

2 Answers: At 0.5 seconds and at 2.5 seconds

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Work Shown:

Use the same idea as problem 10. Instead of y = 0, use y = 120

y = -16x^2 + 48x + 100

120 = -16x^2 + 48x + 100

-16x^2 + 48x + 100 = 120

-16x^2 + 48x + 100-120 = 0

-16x^2 + 48x - 20 = 0

-4(4x^2 - 12x + 5) = 0

4x^2 - 12x + 5 = 0

Now apply the quadratic formula

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-12)\pm\sqrt{(-12)^2-4(4)(5)}}{2(4)}\\\\x = \frac{12\pm\sqrt{64}}{8}\\\\x = \frac{12\pm8}{8}\\\\x = \frac{12+8}{8}\ \text{ or } \ x = \frac{12-8}{8}\\\\x = \frac{20}{8}\ \text{ or } \ x = \frac{4}{8}\\\\x = 2.5\ \text{ or } \ x = 0.5\\\\$

There are two moments in time when the ball is 120 feet in the air. Those two moments are at x = 0.5 seconds and at x = 2.5 seconds.

At x = 0.5, the ball is going upward. At x = 2.5, the ball is coming back down.

Answer from: Quest

24 units

step-by-step explanation:

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Answer from: Quest

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