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Homework Help: Calc I: Another questions

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data
    If a ball is thrown vertically upward with a velocity of 80 ft/sec,
    then its height after t seconds is given by
    s = 80t - 16t2
    . What is the velocity of the ball
    when it is 96 feet above the ground on its way up?


    2. Relevant equations



    3. The attempt at a solution
    I have found my way to s'(t) = -16h -32t + 80
    which I believe to be right. But now I'm not sure what to do. Do I need to find the second derivative?
     
  2. jcsd
  3. Jan 18, 2010 #2
    Where does 16h come from in your equation for s dot? You just need to solve for the time when the ball is at 96 feet.
     
  4. Jan 18, 2010 #3
    Where did you get the -16h term in s'(t)?
     
  5. Jan 18, 2010 #4
    My intermediate step is: (-16t^2 - 32th - 16h^2 + 80t + 80h + 16t^2 - 80t) / h

    Then I cancel out like terms, factor out the h and cancel it. Then I'm left with s'(t) = -16h -32t + 80.

    Where have I gone wrong?
     
  6. Jan 18, 2010 #5
    You forgot that the derivative is the limit as h approaches 0.
     
  7. Jan 18, 2010 #6
    You left out the most important part of your limit definition!
     
  8. Jan 18, 2010 #7
    Argh, it's these little things that kill me in math!

    So the h term becomes 0, so:

    s'(t) = 80 - 32t

    But I'm still unclear how I get the velocity at 96 feet. I recognize that the velocity is the slope of the tangent line but... Oh wait 32 would be the slope of that line, correct? So the answer is -32??
     
  9. Jan 18, 2010 #8
    First you have to solve the position equation for t, then find the velocity at that t.
     
  10. Jan 18, 2010 #9

    Mark44

    Staff: Mentor

    Use the formula for s(t) to find the time t when s(t) is equal to 96. Then substitute that time into your formula for s'(t) - the velocity.

    For example (and these aren't the right numbers) suppose you found that s(t) = 96 for t = 12 sec. Then you would evaluate s'(12) to get the instantaneous velocity at t = 12 sec.
     
  11. Jan 18, 2010 #10
    Not getting anywhere with that, sorry I need another hint.

    All I've been able to come out with are answers that don't make sense.

    By the way, it's multiple choice, and the possible answers are: 80, -80, 32, 16, -16 feet/sec
     
  12. Jan 18, 2010 #11

    Mark44

    Staff: Mentor

    Show us what you're trying to do and we'll set you straight.
     
  13. Jan 18, 2010 #12
    OK.

    96 = 80t - 16t^2
    96 / 16 = (80t -16t^2) / 16
    6 = 5t - t^2
    y = -t^2 + 5t +6
    y = -1(t + 2)(t + 3)

    soooo t = -2 or -3?

    I feel like Im going about this the wrong way... Plugging either of those in gets a number above 100.
     
  14. Jan 18, 2010 #13

    Mark44

    Staff: Mentor

    So far, so good, but you're making this harder than it should be. I would have written 16t^2 - 80t + 96 = 0 as my 2nd equation here. Then I would divide by 16 on both sides to get
    t^2 - 5t + 6 = 0.
    This factors to (t -3)(t - 2) = 0, so t = 2 or t = 3.
    Where did y come from?
    From my work above, the times when the ball is at 96 ft are t = 2 and t = 3. Use your formula for velocity (s'(t)) to find the velocity at these two times. You should find that the two velocity values are numerically equal but opposite in sign.
     
  15. Jan 18, 2010 #14
    Right you are, so I get 16 on the way up or -16 on the way down. Thanks for your help!
     
  16. Jan 18, 2010 #15

    Mark44

    Staff: Mentor

    Be sure to include units - ft/sec.
     
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