a toy rocket is shot vertically into the air from a 8 - foot - tall launching pad with an initial velocity…

a toy rocket is shot vertically into the air from a 8 - foot - tall launching pad with an initial velocity of 48 feet per second. suppose the height of the rocket in feet t seconds after being launched can be modeled by the function h(t)=-16t² + v₀t + h₀, where v₀ is the initial velocity of the rocket and h₀ is the initial height of the rocket. how long will it take for the rocket to reach its maximum height? what is the maximum height? the rocket will reach its maximum height in second(s).
Answer
Explanation:
Step1: Identify the function coefficients
The height - function is $h(t)=-16t^{2}+48t + 8$, where $a=-16$, $b = 48$, $c = 8$.
Step2: Find the time to reach maximum height
The time $t$ at which the rocket reaches its maximum height for a quadratic function $y = ax^{2}+bx + c$ is given by $t=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 48$ into the formula: $t=-\frac{48}{2\times(-16)}=\frac{48}{32}=\frac{3}{2}=1.5$ seconds.
Step3: Find the maximum height
Substitute $t = 1.5$ into the height - function $h(t)=-16t^{2}+48t + 8$. $h(1.5)=-16\times(1.5)^{2}+48\times1.5 + 8$. First, calculate $-16\times(1.5)^{2}=-16\times2.25=-36$. Then, calculate $48\times1.5 = 72$. So, $h(1.5)=-36 + 72+8=44$ feet.
Answer:
The rocket will reach its maximum height in $1.5$ second(s). The maximum height is $44$ feet.