A ball is thrown horizontally from the top of a tower @180m high with a speed of 20m/s how much later must another ball be projected at an angle of 15° to the horizontal from a point on the ground 20√85m away from the top of the tower if the two balls are to meet at the same point
that depends on how fast the 2nd ball is launched.
If it is sent up at time t=a, then
ball 1.
y = 180-4.9t^2
x = 20t
ball 2:
y = v sin15° (t-a) - 4.9(t-a)^2
x = 20√85 + v cos15° (t-a)
Without knowing v, there is no way to get a final number for a.