Question

a ball is thrown upward from a height of 5 feet with an initial velocity of 48 feet per second. its height, h(t) in feet after (t) seconds is given by the quadratic function h(t)= - 16t² + 48t + 5. find the maximum height the ball reaches and the time it takes to reach that height.
tsi quadratic word problem #1
41 feet
-16.5 feet
48 feet
5 feet

Explanation:

Step1: Identify coefficients

The quadratic function is $h(t)=-16t^{2}+48t + 5$, where $a=-16$, $b = 48$, $c = 5$.

Step2: Find time for maximum height

The time $t$ at which the ball reaches its maximum - height is given by the formula $t=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 48$ into the formula: $t=-\frac{48}{2\times(-16)}=\frac{48}{32}=\frac{3}{2}=1.5$ seconds.

Step3: Find maximum height

Substitute $t = 1.5$ into the function $h(t)=-16t^{2}+48t + 5$.
$h(1.5)=-16\times(1.5)^{2}+48\times1.5 + 5$.
First, calculate $-16\times(1.5)^{2}=-16\times2.25=-36$.
Then, calculate $48\times1.5 = 72$.
So, $h(1.5)=-36 + 72+5=41$ feet.

Answer:

The time it takes to reach the maximum height is $1.5$ seconds and the maximum height is $41$ feet.