Let the launch point be the origin of our coordinate system, so the initial position is $\vec{r}_0 = \vec{0}$. The acceleration due to gravity is a constant vector, $\vec{g}$. The initial velocity vector, $\vec{v}_0$, can point in any direction but has a fixed magnitude, $|\vec{v}_0| = v$.

The position vector $\vec{r}$ of any ball at time $t$ is given by the standard kinematic equation:

$$\vec{r} = \vec{v}_0 t + \frac{1}{2}\vec{g}t^2$$

To find the surface containing all possible positions $\vec{r}$, we can rearrange this equation to isolate the initial velocity term:

$$\vec{v}_0 t = \vec{r} - \frac{1}{2}\vec{g}t^2$$

Now, we take the magnitude squared of both sides. Since the initial speed $v$ is the same for all balls, the left side of the equation will evaluate to a constant value.

$$|\vec{v}_0 t|^2 = \left| \vec{r} - \frac{1}{2}\vec{g}t^2 \right|^2$$ $$(|\vec{v}_0|t)^2 = \left| \vec{r} - \frac{1}{2}\vec{g}t^2 \right|^2$$ $$(vt)^2 = \left| \vec{r} - \frac{1}{2}\vec{g}t^2 \right|^2$$

This equation has the form $|\vec{r} - \vec{c}|^2 = R^2$, which is the vector equation of a sphere. By comparing the terms, we can identify the center and radius of this sphere:

• The center of the sphere is at the position vector $\vec{c} = \frac{1}{2}\vec{g}t^2$. This point is a distance $\frac{1}{2}gt^2$ directly below the launch point.
• The radius of the sphere is $R = vt$.

Thus, at any time $t$, all the balls lie on the surface of a sphere that is expanding and falling.