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The height of a ball above the ground as a function of time is given by the functionℎ()=−32^2+8+3where h is the height of the ball in feet and t is the time in seconds. What is the maximum height of the ball? Round to the nearest tenth (one place past the decimal)

Respuesta :

ANSWER

[tex]3.5ft[/tex]

EXPLANATION

The height of the ball is given by the function:

[tex]h(t)=-32t^2+8t+3[/tex]

To find the maximum height of the function, we have to find the value of h(t) when t is:

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

where b = coefficient of t = 8

a = coefficient of t² = -32

Therefore, we have:

[tex]\begin{gathered} t=-\frac{8}{2(-32)}=-\frac{8}{-64} \\ t=0.125 \end{gathered}[/tex]

Now, find h(0.125):

[tex]\begin{gathered} h(0.125)=-32(0.125)^2+8(0.125)+3 \\ h=-0.5+1+3 \\ h=3.5ft \end{gathered}[/tex]

That is the maximum height of the ball.

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