Question

a cannonball is fired directly upward with an initial velocity of 380 feet per second. its height above the ground at time t can be modeled with the equation h = -16t² + 380t. how high does the cannonball travel before it begins to fall back to the ground? round your answer to two decimal places if needed.

Explanation:

Step1: Identify the vertex - time formula

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate of the vertex (in our case, the time $t$ at which the cannon - ball reaches its maximum height) is given by $t=-\frac{b}{2a}$. In the height function $h = - 16t^{2}+380t$, $a=-16$ and $b = 380$.

Step2: Calculate the time $t$ at maximum height

Substitute $a=-16$ and $b = 380$ into the formula $t=-\frac{b}{2a}$.
$t=-\frac{380}{2\times(-16)}=\frac{380}{32}=\frac{95}{8}=11.875$ seconds.

Step3: Calculate the maximum height

Substitute $t = 11.875$ into the height function $h=-16t^{2}+380t$.
$h=-16\times(11.875)^{2}+380\times11.875$.
First, calculate $(11.875)^{2}=141.015625$.
Then, $-16\times(11.875)^{2}=-16\times141.015625=-2256.25$.
And $380\times11.875 = 4512.5$.
$h=-2256.25 + 4512.5=2256.25$ feet.

Answer:

$2256.25$ feet