A model rocket is fired from the ground at time t=0 seconds. Its
height above the ground is given by the following table.
         Time(Seconds)       Height(Feet)
              0                   0
              3                  756
              6                 1224
              9                 1404
             12                 1296
             15                  900
             18                  216
              2
             18.5
a)Find the model h(t),which gives the rocket's height(in feet) as a
function of time(in seconds). Remember units!

b) Find the derivative model. Remember units!

c) Use the model from part a) to complete the table.

d)What is the rocket's instantaneous velocity at t=6 seconds?

e) What is the rocket's maximun height? When does it reach it?

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Request for Question Clarification by omnivorous-ga on 04 Dec 2002 17:56 PST

Fredewq --

The time seems jumbled at the bottom of column 1 -- could you please correct it?
Ah, this makes me nostalgic for physics classes again!

Best regards,

Omnivorous-GA

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Clarification of Question by fredewq-ga on 04 Dec 2002 20:12 PST

Time (Seconds)        Height (Feet)
       
        0                      0

        3                    756

        6                   1224

        9                   1404
    
       12                   1296

       15                    900

       18                    216
 
        2 

       18.5

Hello and thanks for the question!

a) The model for the data is:

Height (in feet) = -16 * (# of seconds)^2 + 300 * (# of seconds)

I used the substitution method to get this result. For an explanation
of this technique, see:
http://www.ucl.ac.uk/Mathematics/geomath/rev/simnb/sim6.html

b) The derivative model is going to be:

y' = -32 * (# of seconds) + 300

c) Ok let's complete the table using the model from part (a):

For 2 seconds:

Height = -16 * 2^2 + 300 * 2

Height = 536 feet

For 18.5 seconds:

Height = -16 * 18.5^2 + 300 * 18.5

Height = 74 feet

d) To calculate the instantaneous velocity we use the derivative
model.

Feet/second = -32 * 6 seconds + 300

108 feet per second is the instantaneous velocity at 6 seconds.

e) For this last part, we can figure out how much seconds passed when
the rocket had an instantaneous velocity of zero. This would be the
highest point in the rocket's trajectory. To do this we use the
derivative model:

0 f/s = -32 * (# of seconds) + 300

# of seconds = 9.375.

Therefore, the rocket reached its highest point after 9.375 seconds

We can now use the model from part (a) to figure out exactly how high
the rocket was:

Height = -16 * 9.375^2 + 300 * 9.375

Height (in feet) = 1406.25

Search Strategy:

polynomials
parabolas
solving simultaneous equations

If you need clarification on any part please ask before rating this
answer.

Best of luck!

-blinkwilliams-ga