Question

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?

Answer 1

Answer: it will take 8.25 seconds to reach maximum height

Step-by-step 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