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Homework Help: Projectile motion proof

  1. Jul 29, 2006 #1
    If m1 is thrown with an angle of [tex]\theta[/tex] from the ground at the same time m2 drops in free motion from the top of a ceiling, prove that
    (1) If m1 aims at m2, as long as the initial velocity is large enough, m1 will hit m2
    (2) The inital velocity V must be larger than [tex][gR/(sin2\theta]^1/2[/tex]

    My calculations:
    for m1:
    1. R=Vcos[tex]\theta[/tex]t
    2. y1=Vsin[tex]\theta[/tex]t-0.5g[tex]t^2[/tex]
    for m2:

    but i'm stuck on what to do next...
  2. jcsd
  3. Jul 29, 2006 #2

    Doc Al

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    Staff: Mentor

    First fix your equation for y2. Like m1, m2 is falling. Hint: Assume the initial position of m2 is x_0, y_0 (where m1 starts at 0,0).
  4. Jul 29, 2006 #3
    After you fix what Doc Al said, you have started correctly. Now think about what the question is asking you.

    When m1 hits m2, what do you know about both of their final positions? What does this tell you about the final (x, y) position? To keep things straight, I would use x instead of R in your first equation, but that's up to you.

    If you get the first part, the second part shouldn't be too difficult. Just remember the trigonometric identity that [tex]2\sin{\theta}\cos{\theta}=\sin{2\theta}[/tex]. Once you get the physics part of it correctly, then the rest of it is math.
  5. Jul 29, 2006 #4
    is r the range of the projectile??
    however your first question is easily answered. both the bodies will have the same y displacement . so the projectile body will always hit the m2
  6. Jul 29, 2006 #5
    That is not a logical way to arrive at the answer. The y-displacement of the two objects does not have to be the same - what matters is that the final y-coordinates of the objects are the same. In some cases, though, this assumption holds true, but those cases are so few that the assumption is better left unassumed.
  7. Jul 30, 2006 #6
    So my mistake is that for m1 and m2 they have different coordinate references, so that's why i can't use the equations for m1 and m2 together... is that right?
  8. Jul 30, 2006 #7


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    Science Advisor

    What coordinates are you using? Did you understand Doc Al's original point: that m2 is falling?
  9. Jul 30, 2006 #8
    The problem is easiest to solve if you put both objects in the same coordinate system.

    Think about the physical significance of the equations that you have derived.
    [tex]x = v\cos{\theta}t[/tex] - What does this mean about the fired projectile's motion? Notice that I have changed the R to an x, so that we can keep the coordinates straight.

    [tex]y_1 = v_0\sin{\theta}t-\frac{1}{2}gt^2[/tex] - What does this mean about the fired projectile? Is it falling or rising, and which way is it accelerating?

    [tex]y_2 = \frac{1}{2}gt^2[/tex] - Compare this equation to your [tex]y_1[/tex] equation. According to what you have derived here, which way is the ball accelerating?

    Check to see that your equations make sense to you before you start solving for the unknowns. In this case, both objects are falling, so their accelerations are downward.
  10. Jul 31, 2006 #9
    [tex]y_2 = -\frac{1}{2}gt^2[/tex]
    so y1=y2
    => Vsin[tex]\theta[/tex]*t-0.5g[tex]t^2[/tex]=[tex]\-frac{1}{2}gt^2[/tex]
    =>Vsin[tex]\theta[/tex]*t=0 ?
  11. Jul 31, 2006 #10
    (Referring to Doc Al's post) For this question, we set the initial position of [tex]m_{1}[/tex] as the origin. In addition, we assume [tex]m_{2}[/tex] starts from the point [tex](x_{0}, y_{0})[/tex].

    When considering the vertical displacement, we will need a reference. In this case, we set it to be the x-axis. Hence, all measurements of vertical displacement will be with respect to this line.

    Since [tex]m_{2}[/tex] already has some vertical displacement to begin with, the equation you need should be [tex]y_2 = y_{0}-\frac{1}{2}gt^2[/tex]

    P/S In the second part of the question, you need to show that the initial velocity V must be bigger than [tex](\frac{gR}{sin2\theta})^{1/2}[/tex]. What is R?
    Last edited: Jul 31, 2006
  12. Aug 1, 2006 #11
    ok! thank you!
    R is the horizontal length from the point m1 is at rest to where m2 falls on the ground...
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