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JimTheFatJam

Solution for initial variables, launch angle and horizontal travel.

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Posted (edited)

Can someone please help? This isn't homework, its just a quick question. Does anyone know the solution to a projectile motion problem in which the variables launch angle and horizontal travel are given? And also theres a catch, the projectile is thrown from a height and lands at a different height. The initial variables for a problem like this are, 1. Launch angle, 2. Horizontal travel, 3. Launch height, 4. Impact height. The only solutions I found online were only for a projectile being thrown, and landing at the same height. Thanks in advance.

Edited by JimTheFatJam
Clarification

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Just change the final height to minus whatever or the initial height to plus whatever and redo the calculation.

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Posted (edited)

For a "free projectile" (one that has an initial force but no continuing thrust), and ignoring air resistance or other friction, the acceleration is the constant -g, the acceleration due to gravity.  Since the gravity, set into an "xy- coordinate system" with positive y upward and positive x to the right, is <0, -g>, the velocity, at any time t, is <vx, vy- gt> where "vx" is the initial velocity in the x-direction and "vy" is the initial velocity in the y-direction.   

Taking the initial speed to be "v" at angle $\theta$ to the horizontal, the initial velocity is $\left< v cos(\theta), v sin(\theta)\right>$ so we have the velocity at time t to be $\left<v cos(\theta), v sin(\theta)- gt\right>$ and, integrating that with respect to time, and taking the initial position to be (0, 0), the position at time t is $\left<v cos(\theta)t, v sin(\theta)t- \frac{1}{2}gt^2\right>.

Now, suppose, after time t, the projectile is at point (w, h).  That is, that the projectile is distance w from the initial point, horizontally, and at height h (both of which may be positive or negative).  Then we must have $v cos(\theta)t= w$ and $v sin(\theta)t- \frac{1}{2}gt^2= h$.  

How we proceed depends upon exactly what the problem is.   If we are given initial speed and angle of the projectile we can immediately calculate w and h.  If we are given w and h and are asked to find the necessary angle and initial speed to achieve that, we can solve the first equation for t, $t= \frac{w}{v cos(\theta)}$ and put that into the second equation to get an equation, $w tan(\theta)- \frac{1}{2}\frac{gw^2}{v^2cos^2(\theta)}$ relating v and $\theta$ (there will be more than one correct answer).

Edited by HallsofIvy

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Posted (edited)

Change your $ sign to [math] for latex. Your dang close  to latexing on this site.

Obviously you close with a [/math] instruction

Example 

[math]\theta[/math]

Also here here for simpler format 

\(\theta\)

Last one was done in same format of the first post here

https://www.scienceforums.net/topic/108127-typesetting-equations-with-latex-updated/

 

 

Edited by Mordred

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