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?

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.

To answer each question, we can use the kinematic equations that relate an object's displacement, velocity, time, and acceleration.

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

vf = vi + at

Here,
vi = 1.2 m/s (initial velocity, downward)
a = 9.81 m/s^2 (acceleration, downward)
t = ? (time, to be determined)
vf = ? (final velocity, to be found)

Since the ball is thrown downwards, the initial velocity is negative (-1.2 m/s), and the acceleration is also negative (-9.81 m/s^2). So, we'll use the negative values in the equation.

Assuming the ball hits the ground after time t, we can set up the equation:

0 = -1.2 m/s - 9.81 m/s^2 * t

Rearranging the equation to solve for t, we get:

t = (-1.2 m/s) / (-9.81 m/s^2)
t ≈ 0.1226 s

Since the time t represents the time it takes for the red ball to hit the ground, we'll use this value to determine the final velocity. Plugging in the values into the equation:

vf = -1.2 m/s - 9.81 m/s^2 * 0.1226 s
vf ≈ -2.201 m/s

Since speed is a scalar quantity, we take the magnitude of the velocity and disregard the negative sign:

Speed of the red ball right before it hits the ground ≈ 2.201 m/s

2) To find the time it takes for the red ball to reach the ground, we have already determined it in the previous step:

Time taken by the red ball to reach the ground ≈ 0.1226 s

3) To find the maximum height the blue ball reaches, we can use the kinematic equation:

vf^2 = vi^2 + 2ad

Here,
vi = 22.9 m/s (initial velocity, upward)
a = 9.81 m/s^2 (acceleration, downward)
d = ? (displacement, to be determined)
vf = 0 m/s (final velocity at the maximum height)

Rearranging the equation to solve for d, we get:

d = (vf^2 - vi^2) / (2a)
d = (0 m/s - (22.9 m/s)^2) / (2 * (-9.81 m/s^2))
d ≈ 26.4 meters

Hence, the maximum height the blue ball reaches is approximately 26.4 meters.

4) To find the height of the blue ball 1.9 seconds after the red ball is thrown, we first need to determine the displacement of the blue ball during this time.

Using the equation:

d = vit + (1/2)at^2

Here,
vi = 22.9 m/s (initial velocity, upward)
a = 9.81 m/s^2 (acceleration, downward)
t = 1.9 s (time)
d = ? (displacement, to be determined)

Plugging in the values, we get:

d = (22.9 m/s)(1.9 s) + (1/2)(-9.81 m/s^2)(1.9 s)^2
d ≈ 25.98 meters

Therefore, the height of the blue ball 1.9 seconds after the red ball is thrown is approximately 25.98 meters.

5) To find the time after the red ball is thrown when both balls are at the same height, we need to determine when the blue ball's height is equal to the red ball's height.

Using the displacement equation:

d = vit + (1/2)at^2

For the red ball, assuming the initial height is 25.0 meters and the displacement is 0 (when it hits the ground):

0 = (1.2 m/s)t + (1/2)(-9.81 m/s^2)t^2

And for the blue ball, assuming the initial height is 0.8 meters:

0.8 m = 22.9 m/s * t + (1/2)(-9.81 m/s^2)t^2

We can set these two equations equal to each other and solve for t:

(1.2 m/s)t + (1/2)(-9.81 m/s^2)t^2 = 0.8 m

Simplifying this quadratic equation, we get:

-4.9t^2 + 1.2t + 0.8 = 0

Solving this equation will give us the time when the two balls are at the same height.