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Initial velocity given distance, height, theta. Projectile motion

  1. Oct 22, 2013 #1
    1. The problem statement, all variables and given/known data
    A baseball is thrown just so that it is able to pass a wall 10m high at a distance that is 1000m away. If it is aimed 30 degrees above the horizontal, what is the initial speed of the baseball?

    2. Relevant equations
    relevant equations shown at the attempt at the solution

    3. The attempt at a solution
    [itex]Vy^2 = Voy^2 + 2ay[/itex]
    [itex]Voy = √(Vy^2-2ay)[/itex]
    [itex]Voy = √(0^2)-2(-9.8)(10)[/itex]
    [itex]Voy = 14m/s[/itex]

    [itex]Sin = Opposite/Hypotenuse[/itex]
    [itex]Hypotenuse = Opposite/Sin[/itex]
    [itex] V = Voy/Sinθ[/itex]
    [itex] V = 14/Sin30°[/itex]
    [itex] V = 28m/s [/itex]

    answer is actually 107
     
  2. jcsd
  3. Oct 22, 2013 #2
    Well, you assumed that the ball passes the wall at the top of its parabolic trajectory.
    But this assumption is not compatible with the given data.
    Use the vertical velocity (as calculated) to find the time to reach the maximum height. Then see if it can travel 1000 m horizontally during the same time. Or what horizontal velocity will be necessary to do so.

    You just use the conditions given: the height must be 10m when the horizontal distance travelled is 1000 m.
     
  4. Oct 22, 2013 #3

    Simon Bridge

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    If v is the initial speed, then you want to find v so that the trajectory passes through point (1000,10)m - given a 30 deg elevation. You forgot the 1000m part. note ##\sin(30)=1/2##, ##\cos(30)=v\sqrt{3/4}##

    good use of LaTeX BTW:
    tip - put a backslash ahead of the trig function so \sin will render as ##\sin## etc.

    ##opposite=o^2p^2site## to write the word "opposite" you have to put it in \text{}
    so it will format as ##\text{opposite}##
     
  5. Oct 22, 2013 #4
    Okay so,
    y = Vot + 1/2gt^2
    0 = 1/2gt^2 + Vot - y
    0 = 4.9t^2 + 14t - 10
    Then I use quadratic formula and I end up with t = 0.688,
    so this is the time it takes for the vertical velocity to reach the height?

    So now I calculate horizontal velocity necessary to travel 1000m during the same time
    Vox = d/t
    Vox = 1000m/0.688s
    Vox = 1453.488m/s

    However I still don't end up with the right answer when I use the pythagorus to solve for the velocity :(
     
  6. Oct 22, 2013 #5
    What do you mean by i forgot the 1000m part, and what was wrong with my initial calculations
     
  7. Oct 23, 2013 #6
    I'm new here and don't know how to use the fancy text which you guys are using, so I've attached a picture of a worked solution.

    As posted before, you cannot just assume that the y velocity is 0 at the time of crossing the wall, I've tried to show why in my drawing.

    I wish I could explain this better because I do not think just giving you the solution will help, but I'll try.


    The ball is being launched with a particular velocity at the start, with an angle of 30 degrees. Only one velocity will satisfy the conditions of the ball being at (1000,10) after a certain amount of time.

    You know that your x has no acceleration, there are many velocities which will get x to 1000.

    However, at the same time, your y value must equal 10.

    So, this leads to solving for an initial velocity which satisfies both equations

    V(0)(t)+0t^2=1000
    V(0)(t)-4.9(t)^2=10

    (since you have no idea what the final y velocity is and therefore cannot use other kinematic equations)

    hope this helps!

    edit: Didn't know posting solutions is illegal, so I took it down.
     
    Last edited: Oct 23, 2013
  8. Oct 23, 2013 #7

    Simon Bridge

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    @twj2944: is is against forum rules to post worked solutions to homework problems.
    If you are in the same class - I guess we can let you off.
     
  9. Oct 23, 2013 #8

    Simon Bridge

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    You didn't use some of the information supplied to you - often a clue that you missed something.

    Well it does not have to reach that height, it has to pass through the height.
    It just has to skim the top of the wall...
    It can do that in three ways:
    (a) just as it reaches it's maximum height - which seems to be what you keep doing.
    ... but the maximum height may be higher than the wall, in which case it can also do it:
    (b) on it's way up to it's maximum height
    (c) on it's way down from it's maximum height

    So what you want is the par of equations that tell you the time T that it reaches a particular x and y position. This will give you two unknowns - T and v.

    You know an expression for y(t) and x(t) right?
    Then, let the time to reach the wall be t=T - leave T an unknown.
    x(T)=1000m = d
    y(T)10m = h

    write out the two expressions in terms of T.

    Solve the simultaneous equations.
     
    Last edited: Oct 23, 2013
  10. Oct 23, 2013 #9
    I think you misunderstood what I said. This was just to convince yourself that your proposed solution does not satisfy the given data. And you did, I hope.

    To find the correct solution just follow what was already suggested by Simon and twj.
     
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