Question

1. a ball is thrown into the air with an upward velocity of 36 ft/s. its height h in feet after t seconds is given by the function h = -16t² + 36t + 9. a = -16, b = 36, c = 9. a. in how many seconds does the ball reach its maximum height? round to the nearest hundredth if necessary. b. what is the balls maximum height? -b/2a = - (36)/2(-16) = 1.125

Explanation:

Step1: Identify coefficients of quadratic function

The height function is $h(t)=-16t^{2}+36t + 9$, where $a=-16$, $b = 36$, $c = 9$. The time $t$ at which the ball reaches its maximum height for a quadratic - function $y = ax^{2}+bx + c$ is given by $t=-\frac{b}{2a}$.

Step2: Calculate time to reach maximum height

Substitute $a=-16$ and $b = 36$ into the formula $t=-\frac{b}{2a}$.
$t=-\frac{36}{2\times(-16)}=\frac{36}{32}=1.125$ seconds.

Step3: Calculate maximum height

Substitute $t = 1.125$ into the height function $h(t)=-16t^{2}+36t + 9$.
$h(1.125)=-16\times(1.125)^{2}+36\times1.125 + 9$.
First, calculate $(1.125)^{2}=1.265625$. Then $-16\times(1.125)^{2}=-16\times1.265625=-20.25$.
$36\times1.125 = 40.5$.
$h(1.125)=-20.25+40.5 + 9=29.25$ feet.

Answer:

a. $1.13$ seconds (rounded to the nearest hundredth)
b. $29.25$ feet