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Planet problem

  1. Apr 7, 2004 #1
    Problem 9. If a ball is thrown vertically upward on the planet Zeno at 100m/s, then it's approximate height in meters t seconds later is given by h(t)=45t-9t^2. a.After how many seconds does the ball hit the ground?
    b. How high does the ball go?
  2. jcsd
  3. Apr 7, 2004 #2
    What is the next value of t such that h(t) is equal to zero?

    Have you studied derivatives? If so, where is the derivative of h(t) equal to zero?

    If not, h(t) describes a parabola. What is the vertex of it?

    And if all else fails, try graphing it.

  4. Apr 8, 2004 #3
    h(t) = 45t - 9t2
    Wouldn't that imply an initial velocity of 45m/s (and an acceleration of 9m/s2 downward), rather than 100m/s? :confused:

    (You can ignore this if you want, mustang, as I don't wish to confuse you and this has nothing to do with the anwer to your problem. :smile:)
  5. Apr 8, 2004 #4
    Assuming planet Zeno is not in some alternate universe where our traditional laws of physics do not apply, then h(t) = y = h(t) = 45t - 9t2 should indicate an initial velocity of 45 m/s and an acceleration of -18 m/s2 (where negative is downward).

    [tex]y = v_0 t + \frac{1}{2}at^2[/tex]

    In the given, the value -9 is equivalent to the expression [tex]\frac{1}{2} a [/tex]

    [tex]-9 = \frac{1}{2}a[/tex]

    Solving for a yields -18 m/s2

    Sorry for the edits. I'm still learning the tex tags.
    Last edited: Apr 8, 2004
  6. Apr 8, 2004 #5
    I suppose that in the equation:
    h(t) = 45t - 9t2
    t = 0 is not the moment in which the ball was thrown. So that does away with my stupid question. :wink:
  7. Apr 8, 2004 #6
    It still implies an initial velocity of 45 in some units. We don't know that those units are m/s, though...

  8. Apr 8, 2004 #7
    We do, since it says "height in meters t seconds later is given by". If the inital velocity was 45mph the equation would be (neglecting acceleration for this purpose):
    h [meters] = 45 [miles]/[hour] * t [seconds]
    That is nonsensical. :tongue:
  9. Apr 8, 2004 #8
    These units are killing me these days. =\ I totally skipped the specification of units for t and h.

  10. Apr 8, 2004 #9
    You look pale, have some more cookies. :smile:
  11. Apr 8, 2004 #10
    That's true, I haven't had any caffeine or chocolate in a while... Must be withdrawal.

  12. Apr 9, 2004 #11
    Cookies sound good. Got milk?
  13. Apr 9, 2004 #12
    So if the acceleration is -18m/s does that mean that the number of seconds does the ball hit the ground is by using synthetic division and then from that use that number for t to put it in the equation to get the height?
  14. Apr 9, 2004 #13
    What does synthetic division have to do with this? We're not dividing polynomials here.

    All you have to do is factor and find the roots of the equation. This is algebra 1 stuff.

  15. Apr 9, 2004 #14
    t=0 or t=5
    So the time is five seconds.

    So then would the value of t=5 in t.
    The height would be 0???
  16. Apr 9, 2004 #15
    Well, naturally. That's when the function is zero and when the ball hits the ground.

    If you want to find the highest point, you have to complete the square. More algebra 1.

  17. Apr 10, 2004 #16
    The height

    SO by completing the square:
    It would be -9t^2+45t=0
    Is this right??
  18. Apr 10, 2004 #17
    Remember that in order to complete the square, the coefficient of the t^2 term must be 1.

    You can actually use your intuition and figure this out. It's a parabola, so it's going to be symmetric. It has a value of 0 at 0 and 5s. Sometime between those two points of time, it attains its maximum. At what time do you suppose it'll be a max?

    The parabola's symmetric... Symmetric...

  19. Apr 10, 2004 #18
    Regarding the height

    Since it stays in the air for 5 seconds midway at 2.5 seconds it reaches it highest height of -9(2.5)^2+45(2.5)=56.25. Right??
  20. Apr 10, 2004 #19

  21. Apr 10, 2004 #20
    A Big Thank You!

    Thanks to all of you guys especially "COOKIEMONSTER."

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