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Derive intial velocity using projectile motion components

  1. May 5, 2009 #1
    Initial Velocity of Projectile
    given:
    - let theta be: B
    -equation for:
    (E1)horizontal component is x = vxt; where vx= v0cosB
    (E2)vertical component is y= y0 +vyt - (1/2)gt(squared); where vy = v0sinB

    using these components an equation of initial velocity{v0} = R{range of projectile}/ square root of (2y/g) is derived

    the problem is i can't figure out the steps on how to come up with this particular equation of initial velocity using the given components

    knowing that according to galileo "an object projected horizontally will reach the ground in the same time as an object is dropped vertically", i came up with these equations
    derived from (E1): t= x/vx (E3)
    i then substituted E3 in E2:
    y = yo + vy (x/vx) - (1/2)g(x/vx)^2

    then, when i plug in the values ( vy = v0sinB, vx= v0cosB), i ended up with this equation:
    y = yo + v0sinB (x/v0cosB) - (1/2)g(x/v0cosB)^2

    i assumed that yo=o, because it was not found in the final equation (initial velocity eq.):
    y= (xv0sinB/v0cosB)- ((gx^2)/2v0^2cos^2B))
    2v0^2cos^2B(y= (xv0sinB/v0cosB)- ((gx^2)/2v0^2cos^2))}2v0^2cos^2B
    y2v0^2cos^2B= (x2v0cosBv0sinB) - gx^2

    i don't know where to go from here...
    please help, it will be greatly appreciated :smile:
     
  2. jcsd
  3. May 5, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    You have Vx = Vo*cosθ

    You have Vy = Vo*Sinθ

    You know that the time to max height is t = Vy/g

    So the total flight time ... up and down is 2*Vy/g

    The range is given by Vx * Total time.

    Substituting for t, you should get Vx*2*Vy/g = Range
     
  4. May 5, 2009 #3
    thank you :smile:

    i also figured out that the equation for
    initial velocity{v0} = R{range of projectile}/ square root of (2yo/g)
    and i also overlooked that the angle was also given in the problem which is 0 degrees:rolleyes:

    steps:
    solve for t:
    y = yo + vy t - (1/2)gt^2

    instead of y0 it should be y=o, so:

    0 = yo + vyt - (1/2)gt^2
    0= yo + vosin0- (1/2)gt^2
    0= yo + (0)- (1/2)gt^2
    yo = (1/2)gt^2
    2yo = gt^2
    2yo / g = t^2
    t = square root of (2yo/g)

    substitution:
    x = vxt
    x = vocos0t
    x = vocos0(square root of (2yo/g)
    x = vo(1)(square root of (2yo/g)
    vo= x / (square root of (2yo/g)

    x is equal to range:
    vo= R / (square root of (2yo/g)



    anyways, thank you so much again for the replying.. :biggrin:
     
  5. May 5, 2009 #4

    LowlyPion

    User Avatar
    Homework Helper

    You mean you are launching it horizontally from a cliff?

    Well that would be different now wouldn't it. My suggestions were taking into account launch and range on a horizontal surface.

    Fortunately you are all over that.
     
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