Question

when two balls are projected horizontally from the same point and velocity, but the time difference for the second ball to projected is just one second after the first one, at what point in the motion will the balls be closest to each other?

Answer 1

The horizontal distance between the balls doesn't change, because they move with the constant speed along X-axis.

So,

$$x = v _ {0} t _ {0}$$ $$v _ {0} - \text{initial velocity}; t _ {0} - 1 \text{s}.$$

Let's find vertical distance between the balls

$$y _ {1} = \frac {g t ^ {2}}{2}$$ $$y _ {1} - \text{distance, which first ball had passed}$$ $$y _ {0} = \frac {g (t - t _ {0}) ^ {2}}{2}$$ $$y _ {0} - \text{distance, which second ball had passed}$$

So, distance between balls along Y-axis is:

$$y = y _ {1} - y _ {0} = \frac {g t ^ {2}}{2} - \frac {g (t - t _ {0}) ^ {2}}{2} = \frac {g t _ {0} (2 t - t _ {0})}{2}$$

Using Pythagorean theorem:

$$d = \sqrt {y ^ {2} + x ^ {2}} = \sqrt {\left(\frac {g t _ {0} (2 t - t _ {0})}{2}\right) ^ {2} + (v _ {0} t _ {0}) ^ {2}}$$

As, you can see $d(t)$ is increasing function, so

$$d _ {m i n} = d (t _ {0}) = \sqrt {\left(\frac {g t _ {0} ^ {2}}{2}\right) ^ {2} + (v _ {0} t _ {0}) ^ {2}}$$

Where $v_{0}$ - initial velocity; $t_0 - 1$ s.