# Thread: Two balls thrown and collide in mid air, produce an expression for the time between

1. ## Two balls thrown and collide in mid air, produce an expression for the time between

Hello!

Basically I am stuck... been pulling my hair out trying to simplify this and come out with a simple expression!
Also not sure what area of the forum to put this in, but seeing as there is nothing too complicated and its all algebra i think this is the right place.

Heres the problem:
Two balls are thrown at angles $\theta_1$, $\theta_2$, at velocity $V_0$ one after the other such that they collide at point B in mid air, determine the time between the throws (analytically).

Heres what i have done so far:
Resolving horizontally gives:

$t_1 = \frac{\cos(\theta_2) t_2}{\cos(\theta_1)}$
and
$t_2 = \frac{\cos(\theta_1) t_1}{\cos(\theta_2)}$

Resolving Vertically gives:

$V_0 \sin(\theta_1) t_1 -\frac{g}{2}t^2_1 = V_0 \sin(\theta_2) (\frac{\cos(\theta_1)t_1}{\cos)\theta_2)}) - \frac{g}{2} (\frac{\cos^2(\theta_1) t^2_1}{\cos^2(\theta_2)})$

Sub in the expression for $t_2$ to find the eqn in terms of $t_1$ then my plan is to simplify and do the same for $t_2$ until finally i can do:

$\Delta{t} = t_1 - t_2$ where $\Delta{t}$ is the answer to the problem.

Okay, so heres where i am stuck:
Trying to simplify the eqn. for $t_1$, im not sure exactly how to do this:

$t_1 =\frac{2V_0(\frac{\sin(\theta_1-\theta_2)}{\cos(\theta_2)})}{g(1-\frac{\cos^2(\theta_1)}{\cos^2(\theta_2)})}$

I have a similar expression for $t_2$.

How do i simplify that?

EDIT: I am guessing there is a trig identity for bottom part which can be written:

$g\frac{cos^2(\theta_2) - cos^2(\theta_1)}{\cos^2(\theta_2)}$

but i cant remember it!

EDIT 2: Perhaps there isnt, so i tried simplifying and got this:

$t_1 = \frac{2V_0\sin(\theta_1-\theta_2)\cos(\theta_2)} {g\cos^2 (\theta_2)-\cos^2(\theta_1)}$

Is that okay?

2. Where did you get $\sin(\theta_1 - \theta_2)$?

I get:

$t_1 = \dfrac{
2V_0\left(\sin(\theta_1) -
\dfrac{\cos(\theta_1)\sin(\theta_2)}{\cos(\theta_2 )\right)}
}
{g\left(
\dfrac{\cos^2(\theta_1)}{\cos^2(\theta_2)}
-1\right)}
}$

Which then gives:

$t_1 = \dfrac{2V_0\cos(\theta_2)\left(\sin(\theta_1)cos(\ theta_2) - \cos(\theta_1)\sin(\theta_2))}{g\left(\cos^2(\thet a_1)- \cos^2(\theta_2) \right)}$

Or you change the previous one to:

$t_1 = \dfrac{2V_0\left(\sin(\theta_1) - \cos(\theta_1)\tan(\theta_2))}
{g\left(\cos^2(\theta_1)sec^2(\theta_2) -1\right)}}$

3. Two balls are thrown at angles $\theta_1$, $\theta_2$, at velocity $V_0$ one after the other such that they collide at point B in mid air, determine the time between the throws (analytically).

You are not asked to calculate $t_{1}$ and $t_{2}$, only the difference between them.

If you take the equation

$V_{0}\sin\theta_{1}t_{1}-(1/2)gt_{1}^{2}=V_{0}\sin\theta_{2}t_{2}-(1/2)gt_{2}^{2},$

and get just the $gt^{2}$ terms on one side, you can isolate $t_{2}-t_{1},$ (by factorising) and then substitute for the ratio of $t_{1}/t_{2}$ on the other side.