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Assignment Goal:
Describe the motion of a rocket using a quadratic model.
Instructions:
Respond to each of the questions. Your solutions must document all of your work.
Scenario: A toy rocket is launched straight upward from a pad 10 ft above ground level with an initial velocity of 80 ft/sec.
Use the general formula for the vertical position (height) of an object moving under the influence of gravity s=-1/2 gt^2+ v_0 t+ s_0 where s0 is initial vertical position, v0 is initial velocity, and s is vertical position of the object.

1. Define a model (equation) to express the vertical position (height) of the rocket s (in meters) above ground level t seconds after launch. Let g = 32 ft/sec2.

Respuesta :

Answer:

Your question is a little unclear if it wants meters or feet because it has everything listed in nicely feet then you say "(in meters)" at the bottom.

If it wants feet, the answer is

[tex]s=-16t^2+80t+10[/tex]

If it wants meters, the answer is

[tex]s=-4.8768t^2+24.384t+3.048[/tex]

Step-by-step explanation:

To solve, use the quadratic equation.

[tex]y=ax^2+bx+c[/tex]

It wants the x-axis to represent time (t) and the y-axis to represent position (s)

[tex]s=at^2+bt+c[/tex]

It gives us a formula that tells us what a, b, and c equal.

[tex]s=(-1/2)gt^2+v_0t+s_0[/tex]    where g is gravity, v_0 is the initial velocity, and s_0 is the initial starting position.

From here, it is just a plug and chug to get the final answer.

[tex]s=(-1/2)(32)t^2+80t+10\\s=-16t^2+80t+10[/tex]

If it wants everything in meters, just convert feet to meters at every step. You can search to find that 1 meter=3.28084 feet.

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