Deriving a Launch Angle Equation - Projectile Motion -

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Homework Help Overview

The discussion revolves around deriving a launch angle equation in the context of projectile motion. The original poster describes a lab experiment where a projectile launcher must hit a moving target that falls due to gravity. The goal is to establish a relationship for the launch angle, θ, using specific variables related to the projectile's motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessity of deriving an equation for the launch angle, with some questioning the assumptions about the timing of the projectile's release and the target's drop. Others suggest that if the bullet is fired at the moment the target is released, the angle may not need to be complex.

Discussion Status

The conversation is ongoing, with various interpretations of the problem being explored. Some participants have offered methodologies for approaching the derivation, while others express confusion about the original poster's specific question and the requirements of the lab.

Contextual Notes

There are constraints regarding the variables that can be used in the equation, specifically excluding certain kinematic variables like initial and final velocities or time. This limitation is a point of contention in the discussion.

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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 apparent 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
 
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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 fundamental implication of the way in which you treat projectile motion problems.
 
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
http://img323.imageshack.us/img323/1853/untitled5po.png
 
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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.
y1=y2
-or-
vsinθ-.5gt²=vsin0-.5gt²

θ=0°


What exactly is the question you're asking...
 
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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
 
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