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Homework Help: Deriving a Launch Angle Equation - Projectile Motion - Please Help

  1. Sep 27, 2005 #1
    Deriving a Launch Angle Equation -- Projectile Motion -- Please Help

    This is physics in the area of projectile motion. My class and I are doing a lab where we have to calculate the launch angle in order to perform a successful lab. The experiment is as follows: the projectile launcher releases a bullet and the goal is to hit the moving target as it falls down (via gravity). In the diagram I provided (in the 2nd attached file), the three circles is the target which in real life will be a thick piece of cardboard.
    Now, prior to conducting the lab, we have to derive a launch angle equation. That is, θ = ? something...
    We can use the 5 kinematics equations. However, as I've written in the attached files, the equation can only consist (but not necessarily all) of dx (range), dy, and h. Vi, Vf, or ∆t cannot be used (acceleration [9.8] can be used).
    As apparant in the first attached file, I have attempted to get to an equation. I started with thinking that, in order for the bullet to hit the target, the vertical distance (dy) of the bullet has to be the same as h-x (where the target will be). Therefore dy= h - x. And ∆dt (distance of target) is what I put for h-x. Hopefully the rest is clear enough to be comprehendable.
    I am not at all sure if what I've done is correct. If it is, I'm not quite sure what to do next.

    So any help or hints would be greatly, greatly appreciated.

    Thanks in advance
    Last edited: Sep 28, 2005
  2. jcsd
  3. Sep 27, 2005 #2
    I was unable to view your attachments, but by the sound of your project, deriving the equation is not necessary. If you aim at an object, and you fire at the instant the object is released, it will hit the object provided its velocity is great enough. This is a fundemental implication of the way in which you treat projectile motion problems.
  4. Sep 27, 2005 #3
    Deriving an equation is necessary... we have to do it. And you can't arbitrarily release the bullet, the angle has to be calculated.

    Here are the images:

    http://img379.imageshack.us/img379/1393/phsy13ag.jpg [Broken]
    http://img323.imageshack.us/img323/1853/untitled5po.png [Broken]
    Last edited by a moderator: May 2, 2017
  5. Sep 28, 2005 #4
  6. Sep 28, 2005 #5
    The vertical component of both the target and bullet are the same, they will fall at the same rate as long as they are released at the same time, so why is there any question to what angle is required to hit the object?

    After t seconds, the bullet and target will have fallen the same distance vertically, the only difference is that the bullet could have moved horizontally closer to the target.

    If the bullet is fired as the target is dropped, an angle of 0 is needed to hit the target. How? Because the equation for the y displacement of an object in relation to the time and vertical component of velocity is the same for both objects, since the component of x velocity and x distance has no bearing on the change in y.

    I'm not entirely sure what the question is. If they're released at the same time, what does he want?

    y=vy(t) -.5gt²
    If both objects are to meet each other, there y distance would be the same, right?
    so, y1=the vertical displacement of the bullet after t seconds
    y2=the vertical displacement of the target after t seconds.
    Since they are to meet, the vertical displacement of the two objects must be equal. The target has no angle about it, it drops straight down.


    What exactly is the question you're asking...
    Last edited: Sep 28, 2005
  7. Sep 29, 2005 #6


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    Homework Helper

    The following methodology almost never fails.

    1. Construct free body diagrams of relevant objects

    2. Write equations of motion of relevant objects in convenient coordinate systems. In this case there should be 3:

    [tex](1): \ y_1 = f_1(\theta,t)[/tex] .....projectile
    [tex](2): \ x_1 = f_2(\theta,t)[/tex] .....projectile
    [tex](3): \ y_2 = f_3(t)[/tex] .....target

    3. Articulate what you want to know/do. In this case
    [tex]- \ eliminate \ t \ from \ the \ equations[/tex]
    [tex]- \ find \ \theta \ needed \ for \ the \ projectile \ to \ hit \ the \ target[/tex]

    4. Use what you know to get what you want
    - one of the 3 equations allows t to be elimiated from the other equations
    - the other 2 should relate y1, y2, and theta. Be careful how you relate y1 and y2 to each other. Net result is that you should be able solve theta in terms of known parameters
    Last edited: Sep 29, 2005
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