Question

score on last try: 0 of 1 pts. see details for more. at least one scored part is incorrect. jump to first changable incorrect part. get a similar question you can retry this question below a projectile is fired into the air from the ground with an initial velocity of 80 feet per second. the function $h(t)=-16t^{2}+80t$ can be used to determine the height, $h$, of the projectile at $t$ seconds. determine the practical domain and range of the function using a graphing calculator. (round your answers to 2 decimals if necessary) domain: range: question help: message instructor

Explanation:

Step1: Find when the projectile hits the ground

Set $h(t)=0$, so $-16t^{2}+80t = 0$. Factor out $-16t$: $-16t(t - 5)=0$. Solving gives $t = 0$ (launch - time) and $t = 5$ (landing - time). The practical domain is the set of all valid input values for $t$. Since time starts at $t = 0$ and ends when the projectile hits the ground, the domain is $0\leq t\leq5.00$.

Step2: Find the maximum of the function

The function $h(t)=-16t^{2}+80t$ is a quadratic function in the form $y = ax^{2}+bx + c$ with $a=-16$, $b = 80$, and $c = 0$. The $t$ - value of the vertex of a quadratic function $y=ax^{2}+bx + c$ is $t=-\frac{b}{2a}$. Substituting $a=-16$ and $b = 80$ gives $t=-\frac{80}{2\times(-16)}=\frac{80}{32}=2.5$. Substitute $t = 2.5$ into $h(t)$: $h(2.5)=-16\times(2.5)^{2}+80\times2.5=-16\times6.25 + 200=-100 + 200=100$. The minimum height is $h(t)=0$ (at launch and landing). So the range is $0\leq h(t)\leq100.00$.

Answer:

Domain: $0\leq t\leq5.00$
Range: $0\leq h(t)\leq100.00$