Question

A blue ball is thrown upward with an initial speed of 21.4 m/s,
from a height of 0.9 meters above the ground. 2.6 seconds after the
blue ball is thrown, a red ball is thrown down with an initial
speed of 9.2 m/s from a height of 25.3 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/s^{2}.

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

Answer #1

**let after t time the two are at same
height**

**for blue ball**

**y1-0.9 = v1y*t + 0.5*g*t^2**

**y1 - 0.9 = (21.4*t) -
(0.5*9.8*t^2)**

**y1 - 0.9 = 21.4t - 4.9t^2**

**y1 = 0.9 + 21.4t -
4.9t^2.............(1)**

**for red ball**

**time = t-2.6**

**25.3 - y1 = v2*(t-2.6) +
0.5*g*(t-2.6)^2**

**25.3 - y1 = 9.2*(t-2.6) + 4.9*(t^2 + 6.76 -
5.2t)**

**25.3 - y1 = 9.2t - 23.92 + 4.9t^2 + 33.124 -
25.48t**

**25.3 - y1 = 4.9t^2 - 16.28t + 9.204**

**y1 = 16.096 - 4.9t^2 + 16.28t
.........(2)**

**0.9 + 21.4t - 4.9t^2 = 16.096 - 4.9t^2 +
16.28t**

**(21.4-16.28)*t = 16.096-0.9**

**t = 2.967 s <--------answer**

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