Question

a toy rocket is shot vertically into the air from a launching pad 9 feet above the ground with an initial velocity of 160 feet per second. the height h, in feet, of the rocket above the ground at t seconds after launch is given by the function h(t)= - 16t² + 160t + 9. how long will it take the rocket to reach its maximum height? what is the maximum height? the rocket reaches its maximum height at second(s) after launch. (simplify your answer)

Explanation:

Step1: Identify the function type

The height function $h(t)=-16t^{2}+160t + 9$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 160$, $c = 9$.

Step2: Use the formula for the vertex of a quadratic

The $t$-coordinate of the vertex of a quadratic function $y=ax^{2}+bx + c$ is given by $t=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 160$ into the formula.
$t=-\frac{160}{2\times(-16)}$

Step3: Simplify the expression

$t=\frac{160}{32}=5$

Step4: Find the maximum height

Substitute $t = 5$ into the height - function $h(t)=-16t^{2}+160t + 9$.
$h(5)=-16\times5^{2}+160\times5 + 9$
$=-16\times25+800 + 9$
$=-400+800 + 9$
$=409$

Answer:

The rocket reaches its maximum height at 5 seconds after launch. The maximum height is 409 feet.