Yes.

Let the origin be the point where the balls are thrown upward and set a +y direction upward.

The first ball has the position equation:

y = y0 + v0*t + (1/2)at^2

y = 48t - 16t^2

The second ball has a position equation:

y = y0 + v0*(t - 1) + (1/2)a(t - 1)^2 <-- Since it is launched 1 s later.

y = 24(t - 1) - 16(t - 1)^2

y = -16t^2 + 56t - 40

When do these two y values equal each other? When

48t - 16t^2 = -16t^2 + 56t - 40

-8t = -40

t =5 s

-Dan