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So say you have a vertical mast of height $32$m, and then say a particle P is projected horizontally from the top of the mast at $18$ m/s. Then say that a particle Q is projected at $30$ m/s from the bottom of the mast at an angle x above the horizontal. Given that the particles collide and that $\cos(x) = \frac35$. Is there a way to find the time taken for the collision from the moment they were projected?

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The laws of motion are \begin{align} x_1(t)&=v_1^0t\\ y_1(t)&=h-\frac{1}{2}gt^2\\ x_2(t)&=v_2^0\cos(x)t\\ y_2(t)&=v_2^0\sin(x)t-\frac{1}{2}gt^2\\ \end{align} We can see that $x_1(t) = x_2(t)$ is an identity.
Equating $$ y_1(t)=y_2(t) $$ we get $$ t=\frac{h}{v_2^0\sin(x)}=\frac{32m}{30m/s\sqrt{1-\left(\frac{3}{5}\right)^2}}=1.33s $$ A graphic representation

enter image description here

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  • $\begingroup$ I am unfamiliar with the laws of motion $\endgroup$ – Benny Jul 15 '18 at 18:31
  • $\begingroup$ I don't quite understand the second equation what does y1 actually mean? $\endgroup$ – Benny Jul 15 '18 at 18:54
  • $\begingroup$ I gathered that x1 and x2 are the horizontal distances, is y1 and y2 vertical distances, if so shouldn't y1 be 32m? Why isn't y1 the same as h? $\endgroup$ – Benny Jul 15 '18 at 18:56
  • $\begingroup$ @Benny, the first particle don't go straight, but decrease its quote, due to gravity, see the graphic representation I added. $\endgroup$ – enzotib Jul 15 '18 at 20:18
  • $\begingroup$ @Benny, if you were not familiar with any of the material in this answer before, you weren't prepared to answer questions like this. Depending on where you got the question from, it looks like you have some material to learn and/or review, for which you should consult an instructor if you can. $\endgroup$ – David K Jul 15 '18 at 21:06
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I suppose the idea is that both particles follow parabolic trajectories, that is, we assume the "standard" gravitational acceleration, no air resistance, use the flat-non-rotating-Earth approximation, etc.

The thing about gravity is that it affects both particles identically. That is, if one particle is deflected downward $1$ meter at time $t,$ then the other particle also is deflected downward $1$ meter. So with respect to whether the particles collide, and how long it takes, the effects of gravity cancel out (unless they cause some other event to occur before the particles can collide, such as one particle collides with the ground).

So first solve the problem assuming no forces apply at all after the instant when the particles are projected. So, aside from all the other simplifying assumptions, assume there is no gravity. (You can think of this as solving the problem in an accelerating frame of reference that happens to be in free fall.)

Compute $\sin(x)$ from the given value of $\cos(x).$ From this you get the vertical velocity of the second particle, and you get the time $t_1$ that it takes the second particle to reach the height $32$ meters.

As a sanity check, compute the horizontal positions of both particles at time $t_1.$ If they are not the same, there is an error either in your calculations or in the problem statement.

As another sanity check, you can take the solution for $t_1$ and find where the particles are at that time if you have "standard" gravitational acceleration.

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  • $\begingroup$ why would I need figure out the time it takes the particle to get to 32 m, if particle P follows a parabolic path? $\endgroup$ – Benny Jul 15 '18 at 18:33
  • $\begingroup$ @Benny In that part of the method you are computing the path without gravity (or in a freely-falling frame of reference; take your pick), and in that case the path is a straight line. In fact, the only reason to even consider gravity is if you're concerned that the particles might hit the ground instead of hitting each other--and at that point you're not trying to find the time of collision, you're merely trying to check an answer you already found to make sure it's correct. $\endgroup$ – David K Jul 15 '18 at 20:59

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