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A weird projectile problem

  1. Feb 20, 2007 #1
    Hi i really do not know how to do this. any help is much appreciated. thanks!

    1. The problem statement, all variables and given/known data

    A projectile is shot out at 45° with respect to the horizon with an initial velocity of Vx = Vy = Vxy. When is the earliest time, t, that the velocity vector makes an angle of 30° with respect to the horizontal?


    2. Relevant equations

    ?

    3. The attempt at a solution

    ok well first i noticed that Vx=Vy=Vxy. then i decided that initially, the horizontal component of velocity is Vxycos45 and the vertical component is Vxysin45. then i thought that where the angle equals 30, the horizontal component will still be Vxycos45 but that the vertical component would be Vxysin30. then i came up with the equations: Vx=Vxycos45 and Vy=Vxysin30t-1/2gt^2. I then tried to set these equations equal to each other to solve for t but i could not get the right answer. i know that the answer is t=Vxy(1-tan30)/g, but i cannot seem to arrive at this answer. please help and tell me if i'm approaching this the right way. thanks a lot
     

    Attached Files:

  2. jcsd
  3. Feb 20, 2007 #2
    Okay, this is just a standard trig problem. since there is no forces affecting the horizontal plane, velocity in the x is a constant. if you wana know when the angle is 30, and you have a constant side. so ( tan(30) ) = ( Vy / Vx ). and the other problem with your equasion. "Vy=Vxysin30t-1/2gt^2" is false. Y=Vxysin30t-1/2gt^2 take note that its equal to 'Y' (the distance) your looking for the ratio of the velocities.
     
  4. Feb 20, 2007 #3
    oh my fault i meant to put Vy=Vxysin30-gt

    i don't know why i put that
     
  5. Feb 20, 2007 #4
    so you have a soln?
     
  6. Feb 20, 2007 #5
    i dont think this part of the data is right...it seems incorrect to me....
    [tex] v(x) = v(y) = v(xy) [/tex]

    if just [tex] v(x) = v(y) [/tex]

    things may be a lot simpler.....
     
  7. Feb 21, 2007 #6
    Well Vx=Vy. Usually the total velocity gets called something different than Vxy. Vxy if total can't be equal to either Vx or Vy. By virtue of 45 degrees, Vx does equal Vy, but nothing more.
     
  8. Feb 21, 2007 #7
    I agree that Vxy can't equal Vx or Vy...however, I do have somewhat of a solution (Kinda sketchy :p):

    We know that wherever this 30 degrees point is, the Vx will be the same as the horizontal so it can be denoted Vy since Vx = Vy. To find the y component, it would be tan30*Vy (Soh Cah toa). Then you know the acceleration is g so we can use:
    g= Dv/Dt ---> t= Dv/g ---> t= (V2-V1)/g ---> t= (Vytan30 - Vy)/g -->
    t= Vy(tan30 -1)/g -----sub Vxy for Vy since they are equal(?) --> t=Vxy(tan30-1)/g
     
  9. Feb 22, 2007 #8
    looks good: the way I solved it was similar;

    Vy(t)=Vy(init)-g*t
    Vx(t)=contant=Vx(init)

    at 30 degrees, Vy(t)/Vx(t)=tan 30=1/2, and dividing the above eqns,

    1/2=1-g*t/Vy or t=Vy/(2*g)
     
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