29.7k views
6 votes
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-example-1

2 Answers

4 votes

Answer:

(I'm not sure if your teacher is nit-picky about significant figures, but here I go...)

8) 1.5 seconds

9) 136 feet

10) 4.415 seconds

11) 2.5 seconds (also 0.5 seconds)

Explanation:

See attached screenshot of Desmos graph showing y(t) = -16t² + 48t + 100, y = 120 and respective points.

I would appreciate it if you could answer the questions please- A ball is thrown upward-example-1
User Gabe Hackebeil
by
8.5k points
4 votes

Problem 8

Answer: 1.5 seconds

----------------

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.

==============================================

Problem 9

Answer: 136 feet

----------------

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.

==============================================

Problem 10

Answer: approximately 4.415 seconds

----------------

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 = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(48)\pm√((48)^2-4(-16)(100)))/(2(-16))\\\\x = (-48\pm√(8704))/(-32)\\\\x \approx (-48\pm93.29523032)/(-32)\\\\x \approx (-48+93.29523032)/(-32)\ \text{ or } \ x \approx (-48-93.29523032)/(-32)\\\\x \approx (45.29523032)/(-32)\ \text{ or } \ x \approx (-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.

==============================================

Problem 11

2 Answers: At 0.5 seconds and at 2.5 seconds

----------------

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 = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(-12)\pm√((-12)^2-4(4)(5)))/(2(4))\\\\x = (12\pm√(64))/(8)\\\\x = (12\pm8)/(8)\\\\x = (12+8)/(8)\ \text{ or } \ x = (12-8)/(8)\\\\x = (20)/(8)\ \text{ or } \ x = (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.

User CherryQu
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories