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Projectile Motion

  1. Jun 17, 2009 #1
    The equation describing the horizontal range of a projectile is

    R = \frac{v_0^2}{g} \sin{2\theta}

    The previous equation states that the projectile wouldn't travel any distance if launched at 0 degrees. Further, the range of a projectile decreases from 45 to 0 degrees, but it seems counterintuitive, since the horizontal distance traveled is described by,

    x = v_0 \cos{\theta_0}t

    Therefore, an angle of 0 would yield maximum x value.

    I'm rather confused.
    Last edited: Jun 17, 2009
  2. jcsd
  3. Jun 17, 2009 #2


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    Hi yitriana! :smile:

    (have a theta: θ and a degree: º :wink:)
    Yes, the range is v0cosθ times t,

    and cosθ is maximum when θ = 0,

    but t is the time in the air, and it isn't constant.

    sin2θ is maximum when 2θ = 90º, ie θ = 45º. :wink:
  4. Jun 17, 2009 #3
    Write out the equations of motion for both vx and vy at time t=0 for initial velocity v0.

    vx0 = v0 cosθ
    vy0 = v0 sinθ

    The maximum height of the projectile is given by
    mgh = (1/2)mvy02; solve for h
    the time the projectile is in the air is given by
    t=2 sqrt(2 h/g)
    The range is
    R=vx0 t

    α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
  5. Jun 17, 2009 #4
    Yes, I see that R is maximized at θ = 45º

    However, if we were to find the x distance of the projectile, wouldn't the distance be maximized when θ= 0, since vx0 = v0 cosθ is greatest?

    How are the x distance and range different when θ = 0º?

    I know they are different when θ != 0º

    It seems logical that if the projectile goes in a straight line with constant velocity, it will travel farther than if the projectile goes in a curved path (at an angle) at constant x velocity.

    So how do the two equations differ when launching a projectile at 0º??
  6. Jun 17, 2009 #5
    The assumption is that friction with the ground will halt the horizontally fired projectile immediately. In other words, [itex]R = 0[/itex] when theta is zero, no matter what the initial velocity is. If your cannon is actually some vertical distance above the ground when it fires, then the formula for range needs some modification. Geometrically, you're looking for the intersection of a line (the ground) and a parabola (the trajectory).
  7. Jun 17, 2009 #6
    Another way to think about it is if [tex] \theta = 0 [/tex] then

    [tex] t = \frac{vsin\theta}{g} = \frac{vsin(0)}{g} = 0 [/tex]

    since the time of flight is 0, then the distance x is 0.

    Btw [tex] t = \frac{vsin\theta}{g} [/tex] comes from doing [tex] v_f = v_0 + gt [/tex] for the y direction, from t = 0, to t = t/2 (aka halfway through). This means [tex] v_f = 0 [/tex] since the ball stops moving in the y direction at that point before starting to come down.
  8. Jun 17, 2009 #7
    Thanks for all your replies!
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