Question

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A ball is thrown into the air with an initial upward velocity of 48 ft/s. It height h in feet after t seconds is given by the function h (t)=-16t^2+48t+4.

What is the ball's maximum height?

Answer 1

Answer:

40 feet (after 1.5 seconds).

Step-by-step explanation:

The height h of the ball after t seconds is modeled by the function:

$h(t)=-16t^2+48t+4$

And we want to determine the ball's maximum height.

Since the given function is a quadratic, the maximum height occurs at the vertex point.

For quadratics, the vertex point is given by the formulas:

$\displaystyle \Big(-\frac{b}{2a},f\Big(-\frac{b}{2a}\Big)\Big)$

In this case, a = -16, b = 48, and c = 4.

Therefore, the t-coordinate at which the vertex occurs is:

$\displaystyle t=-\frac{48}{2(-16)}=\frac{48}{32}=\frac{3}{2}$

So, the maximum height occurs after 1.5 seconds.

Then the maximum height will be:

$\begin{aligned}\displaystyle h\Big(\frac{3}{2}\Big)&=-16\Big(\frac{3}{2}\Big)^2+48\Big(\frac{3}{2}\Big)+4\\\\ &=-16\Big(\frac{9}{4}\Big)+24(3)+4\\\\&=-4(9)+72+4\\\\&=40\text{ feet}\end{aligned}$

So, the maximum of the ball is 40 feet (after 1.5 seconds).