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Given that two projectiles are fired at the same velocity, from the same spot, and land at the same point following different trajectories, how can I find the difference between the two angles of projection and the times taken for the projectiles to land?

EDIT: Not a duplicate since this one involves two projectiles and two different angles rather than trying to find one unique one.

Plato
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3 Answers3

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The equations of motion are $$\left\{\begin{array}{rcl}x(t) &=& v t\cos(\theta)\\y(t)&=&vt\sin(\theta)-\frac{1}{2}gt^2\end{array}\right. $$ hence the projectile travels for a time equal to $\frac{2v\sin\theta}{g}$, landing at a point whose distance from the origin is $$ (v\cos\theta)\cdot\frac{2v\sin\theta}{g} = \frac{v^2}{g}\,\sin(2\theta) $$ and if two different projectiles land at the same point, $\sin(2\theta_1)=\sin(2\theta_2)$ holds, from which $\theta_1=\theta_2$ or $\theta_1+\theta_2=90^\circ$.

Jack D'Aurizio
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Horizantal range = vvsin(2θ)

From this we get that sum of the two angles will have to be 90 degrees.

Time = {2vsin(θ)}/g From this, the ratio of the times will be tanθ.

goalpost
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    Your redaction is very very allusive. You should be more explicit ; for example $\sin(2\theta_1)=\sin(2\theta_2)$ and $\theta_1 \neq 2\theta_2$ implies $2\theta_2=\pi-2\theta_1 \ \iff \ \theta_1+\theta_2=\pi/2$, etc. – Jean Marie Jan 28 '17 at 10:39
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Re: angles. Shreyash seems right. Consider: there's only one possible trajectory for a maximum range shot, 45deg. there are a pair of possibilities for a shot of 0 range, one cannon fires straight up (90 deg) and the other fires along the ground plane (0 deg). Intuitively, the relation should hold for intermediate pairs of ranges: that the sum of the cannon angles will = 90deg. The same relation would hold of the incoming trajectories at the target point.

Aaron Kambeitz
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