# Initial velocity to get from on known position to another?

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1. Sep 26, 2016

### floyd0117

I have a problem where I need to figure out the initial velocity vector $\vec{v_0}$ of a projectile, in order for it to land at the final position $\vec{r_f} = x_f\hat{x} + y_f\hat{y} + z_f\hat{z}$, from initial position $\vec{r_0}$.
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The only knowns in the problem are $\vec{r_0}$ and $\vec{r_f}$. Air resistance is neglected, so the the components of the net force on the projectile are

$m\ddot{x} = 0$

$m\ddot{y} = 0$

$m\ddot{z} = -mg$

So really we can choose any launch angle $\phi$, and find the necessary $|\vec{v}|$, or the other way around, to land us at $\vec{r_f}$. I think it sounds easier to choose a $\phi$ and then find $|\vec{v}|$. So, I examine the limiting cases...
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Let's say $d$ is the the distance between the initial and final positions in the $x$-$y$ plane, that is;

$d = |x_f\hat{x} + y_f\hat{y}|$

and that $h$ is the desired final height, $h = z_f$.

Then the angle $\theta$ measured form the $x$-$y$ plane to a line connecting $(x_0, y_0 ,z_0)$ to $(x_f, y_f, z_f)$ is smiply

$\theta = \arctan{\dfrac{h}{d}}$

So, our limiting cases are:

$\phi \rightarrow \theta; |\vec{v_0}| \rightarrow \infty$

$\phi \rightarrow \dfrac{\pi}{2}; |\vec{v_0}| \rightarrow \infty$
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So I can choose any angle between $\dfrac{\pi}{2}$ and $\theta$, though angles close to those values will necessitate a very large initial velocity. My question is, how do I go from here, to determining $|\vec{v_0}|$? If I choose a $\phi$, how do I find a velocity that will get me to $\vec{r_f}$? It would seem that I need some function of $v_0$ in terms of both $\phi$ (known, after choosing), and $\vec{r_f}$. Am I severely over thinknig this?

Last edited: Sep 26, 2016
2. Sep 26, 2016

### haruspex

You don't go from there. Knowing the range of angles is not much help.
Just suppose the flight time is t and write an expression for where it will be at time t.

3. Sep 26, 2016

### A.T.

The trajectory is a parabola. You know 2 points on it, and the slope (1st deviate) at one of them.