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Projectile Motion (Initial Velocity)

  1. May 4, 2011 #1
    Hi, im trying to find the initial velocity of this question:

    Using a video camera, it is observed that when a ball is kicked from A, it just clears the top of a wall at B as it reaches its maximum height. Knowing the distance from A to the wall is 20m and the wall is 4m high, determine the initial velocity at which the ball was kicked. Neglect size of the ball ( Ans Ua = 23.9m/s).

    So im thinking to find the angle you have to use trig :

    so 20^2 + 4^2 = square root of 416 = 20.4m (hypotenuse)

    using Sin, 4/ 20.4 = Invert Sin 0.196 = 11.3 degrees angle the ball is projected.

    Now i have then angle what else do i need to find in order to find the initial velocity at which the ball was kicked?

    Thanks for any help
  2. jcsd
  3. May 4, 2011 #2

    Doc Al

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    That angle is not the angle of the ball's initial velocity. The ball travels a parabolic trajectory, not a straight line.

    Analyze the vertical motion first. Figure out the vertical component of the initial velocity.
  4. May 4, 2011 #3
    Ah ok, i thought about the ball not traveling in a straight line assuming the angle was wrong at some point.

    Ok so if 'm trying to analyze the vertical component of the initial velocity im going to use equation:

    Sy = Uyt + 0.5(ay)t^2 Or

    Vy^2 = Uy^2 - 2g(Sy)

    The 2nd equation i'm thinking i should use since it has just one unknown (Uy).

    Vy = 0, Sy = 4m

    0 = Uy^2 - 2(-9.81)(4)

    I dont want to go further until i know what i'm doing so far is correct..
  5. May 4, 2011 #4

    Doc Al

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    Almost perfect. You have an extra minus sign in your last equation. g is just the magnitude of acceleration, so g = 9.81, not -9.81. (The acceleration is -g = -(9.81) = -9.81 m/s^2.)
  6. May 4, 2011 #5
    Ok so

    Uy^2 = 2(9.81)(4) + 0

    Uy^2 = square root 78.48

    Uy =8.86 m/s Correct?

    What shall i do now?

    Do i find the time?

    Sy = Uyt + 0.5(9.81)t^

    Sy = 4 + (8.86)t - 4.905t^2

    Sy = 12.86t / 4.905^2

    4 = 12.86 - 4.905

    4 = 7.955

    t = square root 0.5 = 0.7s

    t = 0.7 s .. is this correct so far?
  7. May 4, 2011 #6

    Doc Al

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    Yes. Good.

    Right approach, but you messed up the equation:
    (1) 4 is the final position, not the initial.
    (2) a + bt ≠ (a + b)t

    Even easier: What's the average vertical speed? Use that to find the time.
  8. May 4, 2011 #7
    Average vertical speed..

    Uy / g

    8.86 / 9.81 = 0.9 s?

    (Sorry im getting tired as its nearly 2am)
  9. May 4, 2011 #8
    speed is calculating in m/s.. sorry im confused
  10. May 5, 2011 #9

    Doc Al

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    No. The vertical speed varies uniformly from its initial value of 8.86 m/s to 0 at the highest point. What do you think the average vertical speed would be?
  11. May 5, 2011 #10
    sice it highest point is Sy = 4m

    is it 8.86 / 4 = 2.215 m/s average vertical speed?

    if its not can you provide further help? thanks
  12. May 5, 2011 #11

    Doc Al

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    No. You cannot divide a speed by a distance and expect to end up with a speed.

    Try this analogy: You start out making zero money and, uniformly, little by little, end up making $100/day. Over that time period, what was your average salary per day?

    You may want to review some of the basic formulas of kinematics:
    http://www.physicsclassroom.com/Class/1DKin/U1L6a.cfm" [Broken]

    And here's some good stuff about projectile motion:
    http://www.physicsclassroom.com/Class/vectors/u3l2a.cfm" [Broken]
    Last edited by a moderator: May 5, 2017
  13. May 5, 2011 #12
    u+v / 2 = the average?

    So v = 0

    8.86 / 2 = 4.43 m/s
  14. May 5, 2011 #13

    Doc Al

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    Good! So now that you have the average vertical speed and the vertical distance, figure out the time.
  15. May 5, 2011 #14
    with the analogy:

    Say you made 0$ to 100$ over 7 days .. average income per day is 100 / 7 = 14.3$ per day
  16. May 5, 2011 #15
    Sy = (4.43)t - (4.91)t^2

    4 = (4.43)t - (4.91)

    4 = -0.48?

    4 / 0.48 = square root 8.33

    t = 2.887 s ?
  17. May 5, 2011 #16
    Question: Why does the AVERAGE Initial vertical velocity have to be used instead of just the initial vertical velocity so i can understand why?
  18. May 5, 2011 #17

    Doc Al

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    No. Your income increases from 0/day to 100/day. Like this:
    0/day; 1/day, 2/day, 3/day...... 100/day. Average (assuming uniform increase--just like uniform acceleration) is (0 + 100)/2 = 50/day. Half the time you made less, half the time you made more.

    (But don't worry about that analogy if it doesn't help.)
  19. May 5, 2011 #18

    Doc Al

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    No, using that method you would use the initial velocity, not the average velocity. You'll end up with a quadratic to solve.

    Again, using that method you would use the initial velocity.

    I wanted to show you an easier way to get the time, using Distance = ave speed X time. No quadratics to solve. (Do it both ways and compare the answers.)
  20. May 5, 2011 #19
    Yes i realize using quadratics is a way to find time, i didn't know there was an easier way to solve for t thou.

    as for getting the average vertical initial velocity .. i don't know what this answer goes into which equation in order to solve t.

    Sorry i haven't solved for initial velocity before as this topic (projectile motion) is new to me.

    Easier methods are always better :)

    solving for Ux i would have to find time t?
  21. May 6, 2011 #20

    Doc Al

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    You should solve it both ways to make sure you understand them.

    I gave you the needed equation in my last post--in words. (Another version is in the first link I gave you.)

    You're doing OK. Keep at it!

    Yes. To solve for Ux, you'll first need the time. Which is what you're solving for now.
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