A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h(t) = 60t - 16t^2 . What is the maximum height that the ball will reach?
Do not round your answer.

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Аноним Аноним · Answer 1 · 2019-07-05T20:42:28+02:00

Answer:

56.25 feet.

Step-by-step explanation:

h(t) = 60t - 16t^2

Differentiating to find the velocity:

v(t) = 60 -32t

This equals zero when the ball reaches its maximum height, so

60-32t = 0

t = 60/32 = 1.875 seconds

So the maximum height is h(1.875)

= 60* 1.875 - 16(1.875)^2

= 56.25 feet.

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luisejr77 luisejr77 · Answer 2 · 2019-07-05T20:53:55+02:00

Answer: 56.25 feet.

Step-by-step explanation:

For a Quadratic function in the form [tex]f(x)=ax^2+bx+c[/tex], if [tex]a<0[/tex] then the parabola opens downward.

Rewriting the given function as:

[tex]h(t) = - 16t^2+60t[/tex]

You can identify that [tex]a=-16[/tex]

Since [tex]a<0[/tex] then the parabola opens downward.

Therefore, we need to find the vertex.

Find the x-coordinate of the vertex with this formula:

[tex]x=\frac{-b}{2a}[/tex]

Substitute values:

[tex]x=\frac{-60}{2(-16)}=1.875[/tex]

Substitute the value of "t" into the function to find the height in feet that the ball will reach. Then:

[tex]h(1.875)=- 16(1.875)^2+60(1.875)=56.25ft[/tex]