I want to determine the initial velocity given an angle at which a projectile must be launched in order to hit a target located at a distance $D$ and height $Y$ from the origin.

Following the discussion here I have tried to solve for the velocity and came up with the following equation for the velocity: $$v^2 = \frac{5D^2}{D \sin\theta \cos\theta - Y \cos^2\theta}$$ This was derived from $$ D = vtcos(\theta)$$ and $$Y=vtsin(\theta) - 5(t)^2$$ where I rearranged the equation for $D$ for $t$ and subsituted into the equation for $Y$.

Unfortunately, this seems to fail for values where:

$$Y = D,\theta <= 45$$

Why does my method fail for specific values of $Y$ and $\theta$?


closed as unclear what you're asking by user191954, ZeroTheHero, Jon Custer, glS, stafusa Sep 29 '18 at 12:24

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Can you provide any contex? $\endgroup$ – Jasper Sep 28 '18 at 11:34
  • $\begingroup$ Welcome to Physics SE! Could you maybe add some context, I find it hard to answer without knowing what D,Y, $\theta$ are ans without knowing how you found this expression. $\endgroup$ – Q.Reindeerson Sep 28 '18 at 11:34
  • $\begingroup$ Edited, the first equation was solved for t and substituted into the second equation. @Q.Reindeerson. $\endgroup$ – Christheyankee Sep 28 '18 at 11:37
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    $\begingroup$ @Christheyankee This still lacks some context, what is D, what is Y these type of things :). Also you are new so it is normal that you are still struggling with that but when you make an edit it is better for it to be fitted inside the question, as it had always been here. What do you mean by it fails? For me the problem is more when $\theta <= 45$ if D = Y and they are both positive for instance. To understand why it fails it might be useful to know which physcal system it describes :) $\endgroup$ – Q.Reindeerson Sep 28 '18 at 11:55
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    $\begingroup$ But what are you actually trying to determine? The velocity at some distance from the origin? $\endgroup$ – nluigi Sep 28 '18 at 12:15

Actually, your method fails for $\theta\leq 45$, not $\theta\gt45$.

Mathematically, this is because for your situation the denominator reduces to $tan\theta-1$, which becomes negative for $\theta\leq 45$. This is a problem when subsequently taking the square root.

Physically, consider what happens for your specific values of $Y=D$; the target is located at distance of $\sqrt{2}D$ at exactly an angle of $\theta=45$ from the origin. If you now shoot your projectile at the exact same angle (or less) you will never reach the target because the tiniest bit of gravity will reduce the $y$-velocity such that it will curve under the target.


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