Finding the angle and velocity to hit a target in t seconds

Question

My goal is to compute an angle + velocity combination to hit a target at point $(x,y)$ in exactly $t$ seconds. (Uniform gravity, no drag, no wind)

I know that the general formula for trajectory is:

$x = vt \cos \theta$,

$y = vt \sin \theta - \frac{1}{2} g t^2$.

But the problem is I think I can't solve the equation with only 2 inputs, $x$, $y$ and $t$.

Any ideas would be gladly appreciated, thank you for your brain time.

I believe this is related to projectile motion and should be there in any fundamental physics textbook. — dineshdileep, Jan 09 '13 at 10:56

score 3 · Accepted Answer · answered Jan 09 '13 at 11:25

3

Note that $(v t\cos(\theta))^{2}+(vt\sin(\theta))^{2}=v^{2}t^{2}=x^2+(y+\frac{1}{2}gt^{2})^{2}$ which gives $$v=\frac{\sqrt{x^2+(y+\frac{1}{2}gt^{2})^{2}}}{t}$$ Now it is fairly easy to get $\theta$ from $x=vt\cos(\theta)$.

answered Jan 09 '13 at 11:25

Dominik

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Did you ever convert this to code? – Matt W Jul 16 '13 at 09:33
No. But that should be fairly easy... – Dominik Jul 16 '13 at 16:08
How about this: brackets = y + .5 * g*t ^ 2; equation = (x^2) + brackets^2; v = math.sqrt( equation ) / t; – Matt W Jul 16 '13 at 16:20
I'm just not sure how to get theta from 'x=vt cos(theta)' – Matt W Jul 16 '13 at 16:22

score 0 · Answer 2 · answered Jan 09 '13 at 10:59

From the first equation,

$$ v t = \frac{x}{\cos{\theta}} $$

Plug this into the second equation:

$$ y = x \tan{\theta} - \frac{1}{2} g t^2 $$

Rearrange and solve for $\theta$. Then use the first equation to solve for $v$. Use the fact that

$$ \cos{\tan^{-1}{z}} = \frac{1}{\sqrt{1+z^2}} $$

Finding the angle and velocity to hit a target in t seconds

2 Answers2