Question

A red ball is thrown down with an initial speed of 1.2 m/s from a height of 26 meters above the ground. Then, 0.6 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 23.7 m/s, from a height of 0.9 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.

1)How long after the red ball is thrown are the two balls in the air at the same height?

Answer 1

\[Y _{1}=26+1,2\times t-(9,81)\times(t ^{2})/2 \]

\[Y_{2}=0,9+23,7\times (t-0,6)-(9,81)\times(t-0,6)^{2}/2\]

if t is solved here that would be 1,447 seconds..

Answer 2

of course Y1=Y2 then solve time t