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Velocity of gravity

  1. Sep 30, 2011 #1
    1. A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground.* Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).*

    How high does it rise and how long does it take to get to its highest point?*




    2. Relevant equations



    3. I really don't know where to start. When it says assume a uniform downward acceleration of 10m/s^2, does that mean that for every second, the velocity goes down 10m/s? So after the first second, the velocity is now 5m/s?
     
  2. jcsd
  3. Sep 30, 2011 #2
    The basic equations of motion:
    [itex]x = x_0 + v_0t + \frac{1}{2}*at^2[/itex]
    [itex]v = v_0 + at[/itex]

    Use these.
     
  4. Sep 30, 2011 #3
    How can I use those if I don't have time?

    I also found this equation:

    s = sqrt(2(a)(12)+(15)^2) which gives me 21.5 meters. What is the 21.5 meters? Is that the highest point in the air?
     
  5. Sep 30, 2011 #4
    That is the displacement, if you calculated it correctly. Add the original position to find the height.

    For constant acceleration: vf^2 = v0^2 + 2ad.

    And yeah, I need to learn LaTex.
     
    Last edited: Sep 30, 2011
  6. Sep 30, 2011 #5
    So that would be adding 12 to 21.5, which is 33.5m. So this is the highest point the ball reaches?
     
  7. Sep 30, 2011 #6
    d = (vf^2 - v0^2)/2a = (0^2 - 15^2)/2(-10) = 225/20 = 11.25m.
    y = 11.25m + 12m = 23.25m.
     
  8. Sep 30, 2011 #7
    Oh, I had 10, instead of -10. That's because it's a downward acceleration?
     
  9. Sep 30, 2011 #8
    [itex]v_0 = 15m/s, h_1 = 12m, v_{top} = 0[/itex]
    Lets choose down as the positive way of moving.
    Thus
    [itex]v = -v_0 + at ==> 0 = -v_0 + at ==> t = \frac{v_0}{a}[/itex]
    So the highest point will be
    [itex]h = h_0 + v_0t + \frac{1}{2}at^2 ==> h = h_1 -v_0\frac{v_0}{a} + \frac{1}{2}a(\frac{v_0}{a})^2[/itex]
    Try the rest yourself.
     
  10. Sep 30, 2011 #9
    Ok, so I got that. It takes 1.5 seconds to reach it's maximum height of 23.25 meters.
     
  11. Sep 30, 2011 #10

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The object goes up as long as its velocity is positive. It is coming down once its velocity is negative. It is at its maximum height when its velocity is 0. You find the "t" for that by setting the velocity equal to 0 and solving for t.
     
  12. Sep 30, 2011 #11
    I understood everything up until you say set velocity equal to 0 and solve for t. If t = v/a Then how can you do 0/a? It will always be 0?
     
  13. Sep 30, 2011 #12

    gneill

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    Staff: Mentor

    Strictly speaking, Δt = Δv/a. If you've got some initial velocity vo and the final velocity is 0, then Δv is -vo.
     
  14. Sep 30, 2011 #13
    Oh I see, so what if it asks for this?

    How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?*

    I still never got an answer if when it says downward uniform acceleration of 10m/s^2 means that for every second, the avg velocity goes down by 10m/s?
     
  15. Sep 30, 2011 #14

    gneill

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    Staff: Mentor

    These questions can be answered by choosing appropriate kinematic equations and solving for the desired variable. (The list of general kinematic equations is quite small, and you should probably memorize them).

    For the first case you can use either a conservation of energy approach, or the vf2 = vi2 + 2ad equation you've already seen. For the second case use the position vs time with initial conditions form of the equation of motion. Solve for time.

    That's what the units of acceleration are interpreted to mean: Meters per second, per second :wink:
     
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