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Math Help - Pre-calc question?

  1. #1
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    Pre-calc question?

    Hey!


    A model rocket is launched. The height, in feet, of the rocket h(t) at t (t>=0) seconds after launch is determined by the equation h(t)= (-1/2t)^2 + 15t. The maximum height of the rocket is 112.5 feet.


    (a) Find the number of seconds after launch it takes for the rocket to reach its maximum height. [Show all work]


    (b) For how many seconds will the rocket be at a height of more than 100 ft? [Show all work]


    ----------------------


    ok so its not like i tried it, i did, and let me show what i did.


    first i plugged 112.5 for h(t) and got an answer of t=15 for part a.


    for part b i plugged in 100 for h(t) and got 10 and 20. but i rejected 20 because thats higher than 15 which is the amount of time to reach the maximum height so that makes no sense. so i did 15 - 10 and got 5 for part b.


    am i correct?
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  2. #2
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    Re: Pre-calc question?

    Hello, Hell0!

    A model rocket is launched.
    The height, in feet, of the rocket h(t) at t (t>=0) seconds after launch
    is determined by the equation: . h(t) \:=\: \text{-}\tfrac{1}{2}t^2 + 15t.
    The maximum height of the rocket is 112.5 feet.

    (a) Find the number of seconds after launch it takes for the rocket to reach its maximum height. [Show all work]

    (b) For how many seconds will the rocket be at a height of more than 100 ft? [Show all work]

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Let me show what i did.

    First i plugged 112.5 for h(t) and got an answer of t = 15 for part (a). . Correct!

    For part (b) i plugged in 100 for h(t) and got 10 and 20. . Right!
    But i rejected 20 because thats higher than 15
    which is the amount of time to reach the maximum height. . This is incorrect.

    The rocket is on its way UP aand reaches 100 feet when t = 10 seconds.

    It reaches maximum height when t = 15 seconds.

    Then it is on its way DOWN.
    It is 100 feet above the ground (again) when t = 20 seconds.

    The rocket is above 100 feet from t = 10 to t = 20.
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  3. #3
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    Re: Pre-calc question?

    i dont really understand the going down part
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  4. #4
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    Re: Pre-calc question?

    Quote Originally Posted by Hell0 View Post
    Hey!


    A model rocket is launched. The height, in feet, of the rocket h(t) at t (t>=0) seconds after launch is determined by the equation h(t)= (-1/2t)^2 + 15t. The maximum height of the rocket is 112.5 feet.


    (a) Find the number of seconds after launch it takes for the rocket to reach its maximum height. [Show all work]


    (b) For how many seconds will the rocket be at a height of more than 100 ft? [Show all work]


    ----------------------


    ok so its not like i tried it, i did, and let me show what i did.


    first i plugged 112.5 for h(t) and got an answer of t=15 for part a.


    for part b i plugged in 100 for h(t) and got 10 and 20. but i rejected 20 because thats higher than 15 which is the amount of time to reach the maximum height so that makes no sense. so i did 15 - 10 and got 5 for part b.


    am i correct?
    I would sketch a graph of h(t). Use a graphing calculator if you need help.

    Use that graph to help you figure out how to answer the last question.
    Last edited by phys251; September 1st 2013 at 07:06 PM. Reason: This is a Pre-calc question, not physics.
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