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Old Jan 7th 2017, 09:23 PM   #1
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Two Dimensional Projectile Question

Recently I took on a problem that I was encountering within my robotics team, and it is certainly proving to be over my head.

The question revolves around finding an initial launch angle using nothing but initial velocity, total distance, and final height. Also, I need the projectile to hit the peak of its arch before it reaches the end, as it is a basketball goal-type of a problem.

In this scenario, the total Distance and Initial Velocity are variables that are constantly being input, while the height is always 2.159 meters upwards.

I've been attempting to solve this for a while, but have gotten very little progress on it. Any help would be appreciated. Thank you.
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Old Jan 8th 2017, 06:59 AM   #2
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I presume you know that, with gravity as the only acceleration, -g, starting with initial speed v, at angle $\displaystyle \theta$ to the horizontal, the (x, y) position after time t is given by $\displaystyle x= vcos(\theta) t$ and $\displaystyle y= vsin(\theta) t- (g/2)t^2$.

So if the final height is $\displaystyle y= vsin(\theta) t- (g/2)t^2= H$ and the total distance is $\displaystyle x= vcos(\theta) t= D$, with height, H, distance, D, and initial speed, v, given, those are two equations to solve for t and $\displaystyle \theta$.

I would start by eliminating $\displaystyle \theta$. A clever way to do that is to use the fact that $\displaystyle sin^2(\theta)+ cos^2(\theta)= 1$ for all $\displaystyle \theta$:
From the x equation, $\displaystyle v^2cos^2(\theta)t^2= D^2$.
We can write the y equation $\displaystyle v sin(\theta) t= H+ (g/2)t^2$
so $\displaystyle v^2 sin^2(\theta)= (H+ (g/2)t^2)^2$.

Adding those equations, $\displaystyle v^2 t^2= D^2+ (H+ (g/2)t^2)^2= D^2+ H^2+ gHt^2+ (g^2/4)t^4$ of $\displaystyle (g^2/4)t^4+ (gH- v^2)t^2+ H^2+ D^2= 0$ which is a quadratic equation in $\displaystyle t^2$
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