the height of a toy rocket that is shot in the air with an upward velocity of 48 feet per second can be…

the height of a toy rocket that is shot in the air with an upward velocity of 48 feet per second can be modeled by the function (f(t)=-16t^{2}+48t), where (t) is the time in seconds since the rocket was shot and (f(t)) is the rockets height in feet. what is the maximum height the rocket reaches?\n16 ft\n36 ft\n48 ft\n144 ft
Answer
Explanation:
Step1: Identify the function type
The function $f(t)=-16t^{2}+48t$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 48$, $c = 0$.
Step2: Find the time of maximum height
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 $b = 48$ into the formula: $t=-\frac{48}{2\times(-16)}=\frac{48}{32}=\frac{3}{2}$ seconds.
Step3: Calculate the maximum height
Substitute $t = \frac{3}{2}$ into the function $f(t)=-16t^{2}+48t$. So $f(\frac{3}{2})=-16\times(\frac{3}{2})^{2}+48\times\frac{3}{2}=-16\times\frac{9}{4}+72=- 36 + 72=36$ feet.
Answer:
36 ft