A blue ball is thrown upward with an initial speed of 20.8 m/s, from a height of 0.6 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 10.4 m/s from a height of 24.6 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.
Wondering what the question is? Is it when do they meet? Is it where do they meet?
To analyze the motion of the blue and red balls, we can apply the equations of motion under constant acceleration.
For the blue ball:
1. Find the time it takes for the blue ball to reach its maximum height:
Use the equation: v = u + at, where
- v is the final velocity (0 m/s at maximum height),
- u is the initial velocity (20.8 m/s),
- a is the acceleration (-9.81 m/s^2, as it is acting in the opposite direction to the initial velocity), and
- t is the time we need to find.
Rearranging the equation, we have: t = (v - u) / a.
Substituting the values, we get: t = (0 - 20.8) / -9.81 = 2.12 seconds.
Thus, it takes the blue ball 2.12 seconds to reach its maximum height.
2. Find the maximum height reached by the blue ball:
Use the equation: s = ut + (1/2)at^2, where
- s is the displacement (maximum height above the ground),
- u is the initial velocity (20.8 m/s),
- a is the acceleration (-9.81 m/s^2), and
- t is the time taken to reach maximum height (2.12 seconds).
Substituting the values, we have: s = (20.8 * 2.12) + (0.5 * -9.81 * 2.12^2) = 22.1 meters.
Therefore, the blue ball reaches a maximum height of 22.1 meters above the ground.
For the red ball:
1. Find the time it takes for the red ball to reach the ground:
We can use the equation: s = ut + (1/2)at^2, where
- s is the displacement (negative value, as it is moving downwards),
- u is the initial velocity (10.4 m/s),
- a is the acceleration (9.81 m/s^2, as it is acting in the same direction as the initial velocity), and
- t is the time we need to find.
Substituting the values, we have: -24.6 = (10.4 * t) + (0.5 * 9.81 * t^2).
Rearranging the equation, we get: 0.5 * 9.81 * t^2 + 10.4 * t - 24.6 = 0.
Solving this quadratic equation, we find two values for t: t = 1.71 seconds and t = -3.15 seconds.
Since time cannot be negative, we discard the negative value.
Hence, it takes the red ball approximately 1.71 seconds to reach the ground.