Answer:
t = 0 at the start of the projection
Step-by-step explanation:
To solve this we need to find the distance between the 2 positions at any given time, then solve for the least distance
Let t be the time of the 2nd ball, so t + 1 is the time of the first ball
Let g be the gravitational acceleration, v be the horizontal velocity
the y coordinates of the first and 2nd balls
[tex]y_1 = -g(t+1)^2/2[/tex]
[tex]y_2 = -gt^2/2[/tex]
The x coordinates of the 1st and 2nd balls:
[tex]x_1 = v(t+1)[/tex]
[tex]x_2 = vt[/tex]
The distance between the 2 balls is
[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]
[tex]d = \sqrt{(v(t + 1) - vt)^2 + (-g(t+1)^2/2 - (-gt^2/2))^2}[/tex]
[tex]d = \sqrt{(vt + v - vt)^2 + g/4(t^2 - (t^2 + 2t + 1))^2}[/tex]
[tex]d = \sqrt{v^2 + (g/4)(-2t-1)^2}[/tex]
[tex]d = \sqrt{v^2 + (g/4)(2t+1)^2}[/tex]
As both v and g are constant and cannot be changed, d is minimum when (2t + 1) is minimum, which happens only when t is minimum = 0
