At what time 𝑡meet after the blue ball is thrown are the two balls at the same height in the air?

Question

A blue ball is thrown upward with an initial speed of 22.6 m/s from a height of 0.85 m above the ground. 2.76 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 10.2 m/s from a height of 28.3 m above the ground. The force of gravity due to Earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

Answer 1

We need to write position (vertical) equation for both and make them equal, then solve for the 'time'.

Let's see:

Blue ball:

XB(t) = 0.85 + 22.6t - 0.5 x 9.81 x t²

Red Ball: (because thrown after 2.76s later)

XR(t) = 28.3 + 22.6(t-2.76)- 0.5 x 9.81 x (t-2.76)²

22.6t - 0.5 x 9.81 x t² = 28.3 - 0.85 + 10.2(t-2.76)- 0.5 x 9.81 x (t-2.76)²

22.6t - 4.91t² = 27.45 + 10.2t - 28.15 - 4.91t² - 37.36 + 27.08t

14.68t = 38.06

t =~ 2.6s

I hope I did not make a calculation mistake, please double check B.B