A red ball is thrown down with an initial speed of 1 m/s from a height of 27 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.
How long after the red ball is thrown are the two balls in the air at the same height?
To determine when the two balls are at the same height, we need to compare the heights of the balls at different time intervals. Let's break down the problem into smaller steps:
Step 1: Calculate the time it takes for the red ball to reach the ground.
We know that the initial velocity (u) of the red ball is 1 m/s, the acceleration (a) due to gravity is -9.81 m/s^2 (negative because it's acting in the opposite direction), and the height (h) is 27 meters. Using the kinematic equation:
h = ut + (1/2)at^2
Substituting the given values, we get:
27 = 1t + (1/2)(-9.81)t^2
Rearranging the equation and solving the quadratic equation gives us:
(1/2)(-9.81)t^2 + t - 27 = 0
Using the quadratic formula, t ≈ 3.602 seconds (ignoring the negative value since time can't be negative in this context).
Step 2: Calculate the height of the blue ball after 0.6 seconds.
We know that the initial velocity (u) of the blue ball is 24.5 m/s, the acceleration (a) due to gravity is -9.81 m/s^2, and the time (t) is 0.6 seconds. Again, using the same kinematic equation:
h = ut + (1/2)at^2
Substituting the given values, we get:
h = (24.5)(0.6) + (1/2)(-9.81)(0.6)^2
Simplifying the equation gives us:
h ≈ 8.808 meters
Step 3: Compare the heights of the two balls.
To find the time when the two balls are at the same height, we need to check when the height of the blue ball is equal to the height of the red ball.
Setting the two heights equal to each other, we get:
8.808 = 27 - 1t + (1/2)(-9.81)t^2
Rearranging the equation and solving the quadratic equation gives us:
(1/2)(-9.81)t^2 - t + 27 - 8.808 = 0
Using the quadratic formula, t ≈ 4.133 seconds (ignoring the negative value since time can't be negative in this context).
Therefore, approximately 4.133 seconds after the red ball is thrown, the two balls are at the same height.
To determine the time when the two balls are at the same height, we can analyze their respective positions over time.
Let's first consider the red ball.
Given:
Initial velocity of the red ball, v_red = 1 m/s
Initial height of the red ball, h_red = 27 m
Acceleration due to gravity, g = 9.81 m/s^2
To find the time it takes for the red ball to reach the ground, we can use the kinematic equation:
h = h_red + v_red * t - 0.5 * g * t^2
Setting h = 0 (since the ball reaches the ground), we can solve for the time t:
0 = 27 + 1 * t - 0.5 * 9.81 * t^2
Simplifying the equation, we get:
0.5 * 9.81 * t^2 - t - 27 = 0
This is a quadratic equation. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where a = 0.5 * 9.81, b = -1, and c = -27.
Substituting these values into the formula, we get:
t_red = (-(-1) ± √((-1)^2 - 4 * 0.5 * 9.81 * -27)) / (2 * 0.5 * 9.81)
Simplifying further:
t_red = (1 ± √(1 + 4 * 0.5 * 9.81 * 27)) / (0.5 * 9.81)
Calculating the result:
t_red ≈ 3.915 seconds
Now, let's consider the blue ball.
Given:
Initial velocity of the blue ball, v_blue = 24.5 m/s
Initial height of the blue ball, h_blue = 0.7 m
Acceleration due to gravity, g = 9.81 m/s^2
To find the time it takes for the blue ball to reach the same height, we can use the same kinematic equation:
h = h_blue + v_blue * t - 0.5 * g * t^2
Since the two balls are at the same height, we set h_red = h_blue = h.
0 = 27 + 1 * t - 0.5 * 9.81 * t^2
Simplifying the equation, we get:
0.5 * 9.81 * t^2 - v_blue * t + (h_blue - h_red) = 0.7 * 9.81 * t^2 - 24.5 * t + 0
This is another quadratic equation. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where a = 0.7 * 9.81, b = -24.5, and c = 0.
Substituting these values into the formula, we get:
t_blue = (-(-24.5) ± √((-24.5)^2 - 4 * 0.7 * 9.81 * 0)) / (2 * 0.7 * 9.81)
Simplifying further:
t_blue = (24.5 ± √(24.5^2 - 4 * 0.7 * 9.81 * 0)) / (0.7 * 9.81)
Calculating the result:
t_blue ≈ 2.5 seconds
Therefore, the two balls are in the air at the same height approximately 2.5 seconds after the red ball is thrown.