A blue ball is thrown upward with an initial speed of 22.8 m/s, from a height of 0.8 meters above the ground. 2.8 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 10.9 m/s from a height of 29.3 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.
Incomplete.
To determine various aspects of the motion of the blue and red balls, we can make use of the kinematic equations. These equations relate several parameters of motion, including displacement, time, initial velocity, final velocity, and acceleration.
For the blue ball:
1. Calculate the time it takes for the blue ball to reach its maximum height:
The initial velocity (u) of the blue ball is 22.8 m/s, and the acceleration (a) due to gravity is -9.81 m/s^2 (negative because it acts against the direction of motion). The final velocity at the maximum height is 0 m/s since the ball momentarily stops before falling back down. Using the equation:
v = u + at
0 = 22.8 + (-9.81)t_max
t_max = 22.8/9.81
t_max ≈ 2.33 seconds
2. Calculate the maximum height reached by the blue ball:
Using the equation:
v = u + at
0 = 22.8 + (-9.81) * t_max
0 = 22.8 - 9.81 * 2.33
0 ≈ 22.8 - 22.8667
The height reached by the blue ball above the ground is approximately 0.067 meters.
3. Calculate the total time of flight for the blue ball:
The blue ball was thrown upward, so it will take the same amount of time to reach the maximum height and return to the ground. Therefore, the total time of flight for the blue ball is approximately 2 * 2.33 = 4.66 seconds.
For the red ball:
1. Calculate the time it takes for the red ball to reach the ground:
The initial velocity (u) of the red ball is 10.9 m/s, and the acceleration (a) due to gravity is 9.81 m/s^2 (positive because it acts in the direction of motion). The equation to use is the same as the one for the blue ball since we are only interested in the time it takes to reach the ground.
0 = 10.9 + 9.81 * t_ground
t_ground = -10.9/9.81
t_ground ≈ -1.11 seconds
The negative value implies that the ball took 1.11 seconds to reach the ground after it was thrown (it would have taken about 1.11 seconds to be thrown). Since negative time doesn't make sense in this context, we can ignore this value and consider the absolute value, i.e., approximately 1.11 seconds.
2. Calculate the final velocity of the red ball just before hitting the ground:
Using the equation:
v = u + at
v = 10.9 + 9.81 * t_ground
v ≈ 10.9 + (9.81 * 1.11)
Now you have the necessary information to analyze the motions of the blue and red balls.