Q:
A blue ball is thrown upward with an initial speed of 23 m/s, from a height of 0.5 meters above the ground. 2.8 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6.6 m/s from a height of 28.9 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 blue ball when it reaches its maximum height?
2)How long does it take the blue ball to reach its maximum height?
3)What is the maximum height the blue ball reaches?
4)What is the height of the red ball 3.64 seconds after the blue ball is thrown?
5)How long after the blue ball is thrown are the two balls in the air at the same height?
6) Which statement is true about the blue ball after it has?
a. reached its maximum height and is falling back down?
b. The acceleration is positive and it is speeding up
c. The acceleration is negative and it is speeding up
d. The acceleration is positive and it is slowing down
e. The acceleration is negative and it is slowing down
7) Below is some space to write notes on this problem
here
To solve these problems, we need to use the equations of motion for objects in freefall, which are as follows:
1) v = u + at, where:
v = final velocity
u = initial velocity
a = acceleration
t = time
2) s = ut + 0.5at^2, where:
s = displacement (change in position)
Now, let's solve each of the problems step by step:
1) To find the speed of the blue ball when it reaches its maximum height, we need to find the time it takes to reach that height. We know that the initial velocity is 23 m/s and the acceleration is -9.81 m/s^2. We can use the equation v = u + at to find the time it takes for the velocity to become zero.
v = 0 (at maximum height)
u = 23 m/s
a = -9.81 m/s^2
0 = 23 - 9.81t
Solving for t, we get:
t = 2.35 seconds
Now, we can substitute this time into the equation to find the speed at maximum height:
v = u + at
v = 23 - 9.81 * 2.35
v ≈ 0.03 m/s
Therefore, the speed of the blue ball when it reaches its maximum height is approximately 0.03 m/s.
2) The time it takes for the blue ball to reach its maximum height is already calculated as 2.35 seconds in the previous step.
3) To find the maximum height the blue ball reaches, we can use the equation s = ut + 0.5at^2. We know that the initial height is 0.5 meters and the time taken to reach maximum height is 2.35 seconds.
s = 0.5 + 23 * 2.35 + 0.5 * (-9.81) * (2.35)^2
s ≈ 13.05 meters
Therefore, the maximum height the blue ball reaches is approximately 13.05 meters.
4) To find the height of the red ball 3.64 seconds after the blue ball is thrown, we need to calculate the displacement using the equation s = ut + 0.5at^2. We know that the initial height is 28.9 meters and the time is 3.64 seconds.
s = 28.9 + (0) * 3.64 + 0.5 * (-9.81) * (3.64)^2
s ≈ -18.2 meters
Therefore, the height of the red ball 3.64 seconds after the blue ball is thrown is approximately -18.2 meters. The negative sign indicates that the ball is below the initial height.
5) To find the time when both balls are at the same height, we need to equate their displacements. Let's call this time t2.
For the blue ball:
s_blue = 0.5 + 23 * t2 + 0.5 * (-9.81) * t2^2
For the red ball:
s_red = 28.9 + 6.6 * (t2 - 2.8) + 0.5 * (-9.81) * (t2 - 2.8)^2
Set s_blue = s_red and solve for t2.
Substituting the values and solving the equation will give the time when both balls are at the same height.
6) After the blue ball reaches its maximum height and starts falling back down, it experiences a negative acceleration due to gravity. The statement "e. The acceleration is negative and it is slowing down" is true about the blue ball after it has reached its maximum height and is falling back down.
7) Notes on this problem:
- The blue ball goes up and then comes back down due to the force of gravity.
- The red ball is thrown down from a higher initial height.
- Both balls experience constant acceleration due to gravity, but in opposite directions.
- Pay attention to the signs of the values (positive or negative) to correctly interpret the direction of motion and height.
A blue ball is thrown upward with an initial speed of 23 m/s, from a height of 0.5 meters above the ground. 2.8 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6.6 m/s from a height of 28.9 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 blue ball when it reaches its maximum height?
2)How long does it take the blue ball to reach its maximum height?
3)What is the maximum height the blue ball reaches?
4)What is the height of the red ball 3.64 seconds after the blue ball is thrown?
5)How long after the blue ball is thrown are the two balls in the air at the same height?
6) Which statement is true about the blue ball after it has?
a. reached its maximum height and is falling back down?
b. The acceleration is positive and it is speeding up
c. The acceleration is negative and it is speeding up
d. The acceleration is positive and it is slowing down
e. The acceleration is negative and it is slowing down