Sarah, Heather and Maya are launching a model rocket in the park. They set up their rocket to lauch at an angle and use their phones to record the flight path of the rocket. By estimating heights/times and using some of the math that they learned in Algebra 1, the girls derived a quadratic equation to describe the rocket's path.
The rocket's height can be found using the following equation:
h(t)=-42t^2+253t+2

Where t is the time in seconds and h(t) is the height in feet.

Answer the following questions in the space provided, and submit your work and a diagram showing where each answer is found.

Round answers to two decimal places.

A) Find the maximum height of the rocket.

B) Find the initial height that the rocket was launched from.

C) How long was the rocket in the air?

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Muscardinus Muscardinus · Answer 1 · 2020-09-25T03:06:16+02:00

Step-by-step explanation:

The rocket's path is modeled by the equation as follows :

[tex]h(t)=-42t^2+253t+2[/tex] ....(1)

t is time in seconds and h(t) is the height in feet.

(A) For maximum height, put [tex]\dfrac{dh(t)}{dt}=0[/tex]

So,

[tex]\dfrac{d(-42t^2+253t+2)}{dt}=0\\\\-84t+253=\\\\t=\dfrac{253}{84}\\\\t=3.01\ s[/tex]

Now put t = 3.01 s in equatioon (1)

[tex]h(t)=-42(3.01)^2+253(3.01)+2\\\\h(t)=383\ \text{feet}[/tex]

(B) The general equation of motion as a function of time t is :

[tex]h(t)=-16t^2+ut+h_o[/tex]

u is initial speed.

If we compare the given equation with this general equation, we found that u = 253 m/s

(c) Let for t time the rocket in the air.

[tex]-42t^2+253t+2=0\\\\t = 6.03\ s[/tex]

Therefore, these are the required solutions.