A red ball is thrown down with an initial speed of 1.0 m/s from a height of 27.0 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 24.5 m/s, from a height of 0.7 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 2.0 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?

Which statement is true about the red ball?

The acceleration is positive and it is speeding up

The acceleration is negative and it is speeding up

The acceleration is positive and it is slowing down

The acceleration is negative and it is slowing down

1. V^2 = Vo^2 + 2a*d.

V^2 = 1^2 + 19.62*27 = 530.74
V = 23 m/s.

2. V = Vo + gt. t = (V-Vo)/g.
t = (23-1) / 9.81 = 2.24 s.

3. hmax = ho + (V^2-Vo^2)/2g.
hmax=0.7 + (0-(24.5)^2)/-19.62= 31.3 m.

4. h = ho + Vo*t + 0.5g*t^2.
h=0.7+ 24.5*(2-0.6) -4.9(2-0.6)^2 = 25.39.

To answer these questions, we need to use the equations of motion that describe the motion of the balls under the influence of gravity. Specifically, we can use the following equations:

1. Speed of an object after falling a certain distance h:
v^2 = u^2 + 2gh
where v is the final velocity (speed), u is the initial velocity, g is the acceleration due to gravity, and h is the height fallen.

2. Time taken for an object to fall a certain distance h:
t = sqrt(2h/g)
where t is the time taken, h is the height fallen, and g is the acceleration due to gravity.

3. Maximum height reached by an object when thrown vertically upward:
h = (u^2 / 2g)
where h is the maximum height reached, u is the initial velocity, and g is the acceleration due to gravity.

Now let's solve each question step by step:

1. To find the speed of the red ball right before it hits the ground, we can use the equation mentioned above with h = 27.0 meters (height above the ground), u = 1.0 m/s (initial velocity), and g = 9.81 m/s^2 (acceleration due to gravity):
v^2 = u^2 + 2gh
v^2 = (1.0 m/s)^2 + 2 * 9.81 m/s^2 * 27.0 m
v^2 = 1.0 m^2/s^2 + 529.74 m^2/s^2
v^2 = 530.74 m^2/s^2
v ≈ 23.03 m/s
Therefore, the speed of the red ball right before it hits the ground is approximately 23.03 m/s.

2. To find the time taken for the red ball to reach the ground, we can use the equation mentioned above with h = 27.0 meters (height above the ground) and g = 9.81 m/s^2 (acceleration due to gravity):
t = sqrt(2h/g)
t = sqrt(2 * 27.0 m / 9.81 m/s^2)
t ≈ 2.61 s
Therefore, it takes approximately 2.61 seconds for the red ball to reach the ground.

3. To find the maximum height reached by the blue ball, we can use the equation mentioned above with u = 24.5 m/s (initial velocity) and g = 9.81 m/s^2 (acceleration due to gravity):
h = (u^2 / 2g)
h = (24.5 m/s)^2 / (2 * 9.81 m/s^2)
h ≈ 31.52 m
Therefore, the maximum height reached by the blue ball is approximately 31.52 meters.

4. To find the height of the blue ball 2.0 seconds after the red ball is thrown, we can use the equation of motion for height. We know that the blue ball was thrown upward 0.6 seconds after the red ball, so the time for the blue ball would be 2.0 - 0.6 = 1.4 seconds. Using this time and the equation h = ut + (1/2)gt^2, where u is the initial velocity, g is the acceleration due to gravity, and t is the time:
h = (24.5 m/s)(1.4 s) + (1/2)(-9.81 m/s^2)(1.4 s)^2
h ≈ 15.80 m
Therefore, the height of the blue ball 2.0 seconds after the red ball is thrown is approximately 15.80 meters.

5. To find how long after the red ball is thrown are the two balls in the air at the same height, we need to find the time it takes for the blue ball to reach the same height as the starting height of the red ball. The starting height of the red ball is 27.0 meters. Using the equation of motion for height and substituting h = 27.0 meters, u = 24.5 m/s, and g = 9.81 m/s^2:
27.0 m = (24.5 m/s)t + (1/2)(-9.81 m/s^2)t^2
Rearranging the equation and solving for t using the quadratic formula, we find
t ≈ 2.53 s
Therefore, approximately 2.53 seconds after the red ball is thrown, the two balls are at the same height.

Now let's analyze the statement about the red ball:

"The acceleration is positive and it is speeding up" - This statement is false.
"The acceleration is negative and it is speeding up" - This statement is also false.
"The acceleration is positive and it is slowing down" - This statement is false.
"The acceleration is negative and it is slowing down" - This statement is true.
The red ball is under the influence of gravity, which causes it to accelerate downward with a constant acceleration of 9.81 m/s^2. Since the acceleration is in the opposite direction of the initial velocity (upward), the acceleration is negative. Additionally, the magnitude of the acceleration due to gravity is constant, so the red ball slows down as it falls.