Throwing baseballs

Since gravity is assumed to be the only force on both baseballs, there will be a constant acceleration downwards, and constant velocity horizontally. Thus we can write the two velocities as functions of time, as:

$v_1(t) = (3, 1, -9.8 \times t)$

$v_2(t) = (-4, 2, -9.8 \times t)$

Taking a dot product and asserting that it is zero we can find when their velocities will be perpendicular:

$(3)(-4) + (1)(2) + (-9.8 \times t)^2 = 0$

$(-9.8 \times t)^2 = 10$

$t = \frac{\sqrt{10}}{9.8}$

$t = 0.322$

All that remains is to find the distance between the two at this time. Noticing that they will always be at the same height, this is simply:

$\sqrt{((3)(0.322)-(-4)(0.322))^2+((1)(0.322)-(2)(0.322))^2}$

Which evaluates to:

$2.28 meters$

The velocity vectors can be written as:

$\vec{v_1} = (3,1,-9.8 t)$

$\vec{v_2} = (-4,2,-9.8 t)$

When their velocities become perpendicular the dot product between the vectors is 0:

$(3,1,-9.8 t) \cdot (-4,2,-9.8 t) = 0$

$3\times -4 + 1\times 2 + (-9.8 T) \times (-9.8 T) = 0$

$T = \frac{\sqrt{10}}{9.8} s$

The position vectors can be written as:

$\vec{r_1} = (3t,t,-4.9 t^2)$

$\vec{r_2} = (-4t,2t,-4.9 t^2)$

So the distance between the baseballs is given by:

$\vec{r_{12}} = \vec{r_1}-\vec{r_2} = (7t,-t,0)$

At the time T:

$\vec{r_{12}} = (7T,-T,0)$

$|\vec{r_{12}}| = T \sqrt{50} = \frac{\sqrt{10}}{9.8} \times \sqrt{50} = 2.28 \; m$

Baseball velocities are perpendicular when dot product equals zero. Since both velocities are vector variables in time, this gives us the time at which this happens. Integrate vector velocities with initial condition $S_{i}(t=0) = (0,0,0)$ atop tower to find $S_{i}(t)$ for either baseball. Figure distance between baseballs at time figured above to answer problem.

Integrating acceleration into velocities of both, with either initial (vertical) velocity = 0 gives:

$V_{1}(t) = (3, 1, -9.8t)$

$V_{2}(t) = (-4, 2, -9.8t)$

Zeroing the dot product gives that $t = \sqrt{10}/9.8$ , in seconds.

Integrating again to find position vectors, with initial condition $S_{i}(t=0) = (0,0,0)$ gives

$S_{1}(t) = (3t, t, -\frac{1}{2}(9.8)t^{2})$

$S_{2}(t) = (-4t, 2t, -\frac{1}{2}(9.8)t^{2})$

so, the distance between the baseballs at the time figured above is the square root of $50t^{2}$ , or 2.28 meters .

After t second their velocities are perpendicular. So (3,1,-9.8t)T * (-4,2,-9.8t)T = 0,So t = 0.32268 second.Their position now can be calculated by s = v t - 1/2 g t^2. And their positions are (0.96804,0.32268,-0.5096) and (-1.29072,0.64576, -0.5096). And their distance can be calculated using the formula ((x1 - x2)^2 + (y1 - y2)^2 + (z1 -z2)^2 ) ^0.5. And the result is 2.28 m

Create your two velocity vectors, v1=(3,1,-9.8t) and v2=(-4,2,-9.8t), then take the dot product and let it equal to 0. That will let you know that the vectors are perpendicular and therefor that will give you the time at which they are perpendicular. That time should be 0.323 seconds. After that, integrate each velocity vector to get your displacement vectors, and plug in 0.323 seconds. (d1 = (3t,t,-4.9t^2) and d2=(-4t,2t,-4.9t^2)). Then use the formula for the distance between two points to calculate the distance, which should end up being 2.28m.