185k views
4 votes
Suppose that a baseball is thrown upward with an initial velocity of 132 feet per second ​(90 miles per​ hour) and it is released 4 feet above the ground. Its height h after t seconds is given by hequalsminus16tsquaredplus132tplus4. After how many seconds does the baseball reach a maximum​ height?

1 Answer

3 votes

Answer: it will take 8.25 seconds to reach maximum height

Explanation:

Initial velocity, u of baseball = 132 feet per second. The height of the baseball, h in feet after t seconds is given by the function h = −16t^2 + 132t + 4

The given function is a quadratic equation. If values of height attained is plotted against time, the graph will take the shape of a parabola whose vertex corresponds to the maximum height attained by the baseball

Vertex of the parabola = -b/2a

a = - 16

b = 132

Vertex = - 132/-16× 2 = -132/32

Vertex = 4.125

The maximum height is 4.125 feets

To determine the time it will take to reach the maximum height of 4.125 feets, we will substitute h = 4.125 in the equation

4.125 = −16t^2 + 132t + 4

−16t^2 + 132t - 0.125 = 0

Applying the general formula for quadratic equations,

t = [- b ± √b^2 - (4ac)]/2a

a = -16

b = 132

c = -0.125

t = [- 132 ± √132^2 - 4(-16 × - 0.125)]/2× -16

t = [- 132 ± √17424 - 8)]/-32

t = [- 132 ± √17416]/-32

t = (-132 ± 132)/-32

t = (-132 + 132)/-32 or (-132 -132)/-32

t = 0 or -264/-32

t = 8.25 seconds

User Alex Andronov
by
7.6k points

No related questions found

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