Amember of the local rocketry club launches her latest rocket from a large field. at the moment it's fuels is exhausted, the rocket has a velocity of 240 feet per second and a altitude of 544feet per second. a. write the function that models the height of the rocket. b how high is the rocket at t=0? if it took off from the ground, why is this high at t=0? c. how high is the rocket 5 seconds after the fuel is exhausted? d. what is the maximum height attained by the rocket?

Step-by-step explanation:

A. We don't have a standard model for powered rocket behavior. For a given thrust, mass decreases as fuel burns, so acceleration increases. These interacting variables depend on the percentage of the mass that is fuel, the rate at which fuel burns, and the thrust obtained from a given mass of fuel. Likely, the density of the atmosphere and the drag associated with that are involved, as well. Even if these factors are modeled in a simple way, the solution to the equation(s) involved rarely shows up in an elementary physics or math book.

So, we're left with modeling the ballistic motion after the fuel runs out. Here, we usually ignore air drag (or lift), pretending the entire behavior matches that of an object in a vacuum and a uniform gravitational field on a stationary, flat Earth. The equation for that motion is ...

... height = -1/2·g·t² +v₀·t +h₀ . . . . where v₀ and h₀ are the velocity and height at t=0, the time the fuel runs out.

g is the gravitational constant, usually taken to be 32 ft/s² or 9.8 m/s², depending on the units of the other variables.

We are given v₀ = 240 ft/s and h₀ = 544 ft, so the equation of motion is ...

... h(t) = -16t² +240t +544 . . . . ft, for t in seconds.

B. The problem statement tells us the rocket is 544 ft high when the fuel runs out (at t=0). Our model of rocket behavior begins at t=0, which we define as the time the fuel runs out.

C. For t=5, h(5) = -16·5² +240·5 +544 = 1344 . . . ft

D. The graph shows the maximum height to be 1444 ft at t=7.5 seconds.

Analytically, the vertex is found at t=-b/(2a) = -240/(2·(-16)) = 240/32 = 7.5 . . . seconds. h(7.5) = (-16·7.5 +240)·7.5 +544 = 1444 . . . ft.

answer: 20

step-by-step explanation: subtract the minimum date value from the maximum data value to find the date range. in this casem the date range is 42 - 22 = 20

hope this ! : ) ~zane