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Projectile Motion and Velocity Problem

  1. Oct 3, 2008 #1
    A rock is thrown from the top of a 26 m building at an angle of 51° above the horizontal.

    (a) If the horizontal range of the throw is equal to the height of the building, with what speed was the rock thrown?

    (b) How long is it in the air?

    (c) What is the velocity of the rock just before it strikes the ground?


    Let Vi = initial velocity
    X component = Vi Cos 51
    Y component = Vi Sin 51

    I know that without any height, I can calculate the time
    0 - Vi Sin 51 = -9.81 * T
    T = Vi Sin 51 / 9.81

    Total time in the air = 2 Vi Sin 51 / 9.81

    Horizontal range = Vix * Total time in air
    26 = Vi Cos 51 * 2 Vi Sin 51 / 9.81

    but since there is a height involved, what changes do I need to make?
     
  2. jcsd
  3. Oct 3, 2008 #2

    Hootenanny

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    When the rock is thrown off the top of the building what is the rocks final vertical position relative to the top of the building?
     
  4. Oct 3, 2008 #3
    Im not too sure on the question but if a rock is thrown off the top of the building, the final vertical position is another 26 meters down?
     
  5. Oct 3, 2008 #4

    Hootenanny

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    Correct. So you know the final vertical displacement (-26m), the acceleration, intitial vertical velocity and you want to find the time of flight. Can you think of a kinematic equation with all those variables in?
     
  6. Oct 3, 2008 #5
    It would be
    D = Vi * T + 1/2 A * T^2

    But I don't know Vi vertical.
    Vi Vertical is Vi Sin 51, The Vi is still unknown.
     
  7. Oct 3, 2008 #6

    Hootenanny

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    Correct, but you do know the vertical range. Can you form another equation using this information?

    Don't forget that it should be Vi sin(51) not just Vi
     
  8. Oct 3, 2008 #7
    Isn't the vertical range formula the same for horizontal range except the component differences?

    Vertical range formula is
    Dy = Yo + Viy * T

    but Viy = Vi sin(51) we don't know Vi tho.

    I can't think of any more formula that utilizes these info.
     
  9. Oct 3, 2008 #8

    Gib Z

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    The formula's are in fact not the same - The horizontal component has no acceleration, whilst the vertical component is influenced by Acceleration due to Gravity. You know when it has reached the peak of its vertical motion, the y-component of its velocity is equal to ... ? You also know the displacement in the y- direction and acceleration in the y-direction. You want to find the Initial velocity in the y-direction . Try to remember a formula that relates these variables.
     
  10. Oct 4, 2008 #9
    Ahh, is it
    Vf^2 = Vi^2 + 2 * A * D?

    Vi^2 = 510.12
    Vi = 22.5 m/s

    Hmm, I gave this a little thought, Im not sure if I am correct.
    I have the initial vertical velocity now so I will use

    Vf = Vi + A * T
    where Vi = 22.5 sin 51 and A = 9.81
    Where Vf = 0 to find the time that it hits the highest point
    Where T = 1.8s

    Using the time that hits the highest point I plug it in
    D = Vi * T + 1/2 A * T^2
    to find the distance it traveled
    and D = 47m up

    Since it reached 47m up, it must come down another 47m to its initial position thrown
    which is 94m but since it falls to the ground that is another 26m down from the buildings height
    which D = 120m total.

    Now I use
    D = Vi * T + 1/2 A * T^2
    where D = 120 to find the total time in the air
    T = 3.5s

    Plug that in the horizontal range equation
    D = Vix * T
    120 = Vi Cos 51 * 3.5
    Vi = 12 m/s

    Am I Right?
    I hope so!!

    Can someone help me with this
    (c) What is the velocity of the rock just before it strikes the ground?
     
    Last edited: Oct 4, 2008
  11. Oct 4, 2008 #10

    Hootenanny

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    That would be indeed the correct formula, but where have you got that initial velocity from?
     
  12. Oct 4, 2008 #11
    Im srry I made an edit on my post, could you please look at it.
     
  13. Oct 4, 2008 #12

    Hootenanny

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    You still haven't said how you obtained the initial vertical velocity.
     
  14. Oct 4, 2008 #13
    I use the formula that you deemed correct and plug in the numbers.

    Vf^2 = Vi^2 + 2 * A * D
    Vf = 0 because it is at its highest position
    D = 26m

    to find Vi. which is 22.5 and Initial vertical velocity is 22.5 sin 51.
     
  15. Oct 4, 2008 #14

    Hootenanny

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    The problem is that the velocity is non-zero when the displacement is 26m. When the ball is at y=26, the velocity is equal to the initial velocity.
     
  16. Oct 4, 2008 #15
    Ok now Im confused.
    So the final velocity is not 0 when is at y=26 and is equal to its initial? So how would I approach this?
     
  17. Oct 4, 2008 #16

    Hootenanny

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    Yes, because at y=26m the ball is on top of the building. As I said before, use the equation in your above post and fill in all the variables that you know. You will be left with two unknowns, hence you will need to construct a further equation using the information you are given in the question.
     
  18. Oct 4, 2008 #17
    Oh so what I did before was wrong then.

    Then we are back to square 1 now but even worse, we have 2 unknowns.
    Vf^2 = Vi^2 + 2 * A * D
    Vf^2 = Vi^2 + 2 * -9.81 * 26

    Can you throw me a hint?
     
  19. Oct 4, 2008 #18

    Hootenanny

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    Consider the vertical motion first. Try using this equation that you posted before:
    Fill in all the information you know.
     
  20. Oct 4, 2008 #19
    D = Vi * T + 1/2 A * T^2

    26 = Vi * T + 1/2(9.81) * T^2

    Unknowns are Vi and T
     
  21. Oct 4, 2008 #20

    Hootenanny

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    Close, but don't forget that the final position of the ball is below it's initial position and therefore D=-26m. Further, the acceleration due to gravity is in the negative y direction and therefore a=-9.81 m/s2.

    So you have two unknowns, Vi and T. Can you now write down an equation that governs the horizontal motion?
     
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