Question

if a ball is thrown directly upward with a velocity of 46 ft/s, its height (in feet) after t seconds is given by y = 46t - 16t². what is the maximum height attained by the ball? (round your answer to the nearest whole number.)

Explanation:

Step1: Identify the function type

The height - function $y = 46t-16t^{2}$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 46$, and $c = 0$.

Step2: Find the time $t$ at which the maximum occurs

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate (in our case, the time $t$) of the vertex is given by $t=-\frac{b}{2a}$. Substituting $a=-16$ and $b = 46$ into the formula, we get $t=-\frac{46}{2\times(-16)}=\frac{46}{32}=\frac{23}{16}=1.4375$ seconds.

Step3: Find the maximum height

Substitute $t = \frac{23}{16}$ into the height - function $y = 46t-16t^{2}$.
\[

$$\begin{align*} y&=46\times\frac{23}{16}-16\times(\frac{23}{16})^{2}\\ &=\frac{46\times23}{16}-\frac{16\times23^{2}}{16^{2}}\\ &=\frac{1058}{16}-\frac{16\times529}{256}\\ &=\frac{1058}{16}-\frac{8464}{256}\\ &=\frac{1058\times16}{16\times16}-\frac{8464}{256}\\ &=\frac{16928}{256}-\frac{8464}{256}\\ &=\frac{16928 - 8464}{256}\\ &=\frac{8464}{256}\\ &=\frac{529}{16}\\ &=33.0625\approx33 \end{align*}$$

\]

Answer:

33