A red ball is thrown down with an initial speed of 1.2 m/s from a height of 25.0 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 22.9 m/s, from a height of 0.8 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- What is the speed of the red ball right before it hits the ground? 2) How long does it take the red ball to reach the ground? 3) What is the maximum height the blue ball reaches?4) What is the height of the blue ball 1.9 seconds after the red ball is thrown?5) How long after the red ball is thrown are the two balls in the air at the same height?

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1) To find the speed of the red ball right before it hits the ground, we can use the following equation:

v_f = v_i + g*t

where v_f is the final speed, v_i is the initial speed, g is the acceleration due to gravity, and t is the time it takes for the red ball to reach the ground.

First, we need to find the time it takes for the red ball to reach the ground. Using the equation:

h = v_i * t + (1/2) * g * t^2

We get:

25.0 = 1.2*t + (1/2) * 9.81 * t^2

Solving for t, we get t ≈ 2.25 seconds.

Now, we can find the final speed of the red ball:

v_f = 1.2 + 9.81*2.25 ≈ 23.13 m/s.

2) As calculated above, it takes the red ball 2.25 seconds to reach the ground.

3) To find the maximum height the blue ball reaches, we first need to find the time it takes for the blue ball to reach its maximum height. We can use this equation:

v_f = v_i - g*t

Where v_f at the maximum height is 0. Solving for t, we get:

t ≈ 22.9/9.81 ≈ 2.33 seconds

Now, we can find the maximum height using:

h = v_i * t - (1/2) * g*t^2.

Plugging the values, we get:

h_max = 22.9 * 2.33 - (1/2) * 9.81 * (2.33)^2 ≈ 26.62 meters

Since the blue ball was initially thrown from 0.8 meters above the ground, the total maximum height above the ground is 26.62 + 0.8 = 27.42 meters.

4) To find the height of the blue ball 1.9 seconds after the red ball is thrown, we should first find the time elapsed after the blue ball is thrown which is 1.9 - 0.6 = 1.3 seconds. Now, we can use the equation:

h = v_i * t - (1/2) * g * t^2

Plugging the values:

Blue ball height = 22.9 * 1.3 - (1/2) * 9.81 * (1.3)^2 ≈ 17.82 meters

Since the blue ball was thrown from 0.8 meters above the ground, the total height above the ground is 17.82 + 0.8 = 18.62 meters.

5) To find when the two balls are at the same height, we can set their height equations equal to each other and solve for time.

Red ball height: h_r = 1.2t - (1/2) * 9.81 * t^2 + 25

Blue ball height: h_b = 22.9(t-0.6) - (1/2) * 9.81 * (t-0.6)^2 + 0.8

We need to find t such that h_r = h_b:

1.2t - (1/2) * 9.81 * t^2 + 25 = 22.9(t-0.6) - (1/2) * 9.81 * (t-0.6)^2 + 0.8

Solving this equation, the two balls are at the same height after t ≈ 1.48 seconds after the red ball is thrown.