Ablue ball is thrown upward with an initial speed of 21.6m/s, from a height of 0.8 meters above the grpund. 2.6 seconds after the blue ball is thtown, a red ball is thrown down with an initial spees of 8.3m/s from a height of 26.1 meters above the ground. the force of the gravity due to the earth results in the balls each having a constant downward acceleration of 9.8m/s
1.what is the max. height the blue ball reaches?
2. what is the height of the red ball 3.38 seconds after the ball is thrown
3. how long after the blue ball isbthrown are the two balls in the air at the same height?

When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.

1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²

For the blue ball, v₀ = 21.6 m/s. Substituting the values:
H = (21.6 m/s)²/2(9.81m/s²)
H = 23.78 m
The maximum height reached by the blue ball is 23.78 m + 0.8 = 24.58 m.

2. The distance traveled with any given distance is calculated using the formula:
y = v₀t + 0.5gt²
where y is the distance at any time t
For the red ball, v₀ = 8.3 m/s. Substituting the values:
y = (8.3 m/s)(3.38 s) + 0.5(9.81 m/s²)(3.38 s)²
y = 84.09 m
The red ball would reach 84.09 m after 3.38 seconds.

3. For this, you equate the y values of both balls:

y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(8.3 m/s)t + 0.5(9.81 m/s²)(t²) + 26.1 m = (21.6 m/s)t + 0.5(9.81 m/s²)(t²) + 0.8 m
Solving for t, 
t = 1.9 seconds

Thus, the two balls would be at the same height after 1.9 seconds.