Homework Problem

Let the lower left corner be the origin. The $x$ and $y$ equations for the collision at time $t$ are:

$10 t = d - 10 t \\ 10 + 10t - 5 t^2 = 20 - 5 t^2$

Solving these results in $(t,d) = (1,20)$

We know the initial velocities of points A and B are as follows:

$v_{A,i} = 10\mathbf{i} + 10\mathbf{j}$

$v_{B,i} = 0\mathbf{i} - 10\mathbf{j} = - 10\mathbf{j}$

The only requirement given in the problem is that the points must collide in air, meaning that they will collide in the x direction at the same point.

So, since we know that both particles are traveling toward each other at the same speed, $10 \text{ m/s}$ , then they will travel toward the center of $d$ at the same speed; this means that they must meet at $x = \frac{1}{2}d$ .

Now, the only thing we need to find is the horizontal distance of $d$ .

We know that they'll both reach $x = \frac{1}{2}d$ at the same time, so the equations of motion for each of the particles will have the same value for $t$ , time, and the same value for $x_f$ , which is $\frac{1}{2}d$ .

Using: $x_f = x_i + v_it$ for each of the individual particles, we can solve for the $t$ value they share in common.

For particle A:

$\frac{1}{2}d = 0 + 10t = 10t$

For particle B:

$\frac{1}{2}d = 20 - 10t$

So, setting these equal to each other gives the following equation, from which we can solve for t:

$10t = 20 - 10t$

$\rightarrow 20t = 20$

$\rightarrow t = 1$

So, now that we know that $t=1$ , we can solve for the value of $d$ from either of the equations of motion above.

$\frac{1}{2}d = 10t = 10*1 = 10$

$\rightarrow d = 20$

So, the distance $d$ between the towers is 20 meters .