A blue ball is thrown upward with an initial speed of 21.0 m/s, from a height of 0.8 meters above the ground. 2.6 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 8.3 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.
1) What is the maximum height the blue ball reaches?
2) What is the height of the blue ball 1.9 seconds after the red ball is thrown?
3) How long after the red ball is thrown are the two balls in the air at the same height?
To answer these questions, we need to use basic kinematic equations that describe the motion of objects under constant acceleration.
1) To find the maximum height the blue ball reaches, we can use the equation for vertical displacement:
h = h0 + v0t - (1/2)gt^2
where h is the final height, h0 is the initial height, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity.
In this case, the initial height (h0) is 0.8 meters, the initial velocity (v0) is 21.0 m/s, and the acceleration due to gravity (g) is -9.81 m/s^2 (negative because it acts in the opposite direction of the motion).
Plugging in these values, we have:
h = 0.8 + (21.0)(t) - (1/2)(9.81)(t^2)
To find the maximum height, we need to determine the time at which the ball reaches its highest point. At the highest point, the vertical velocity is 0 (v = v0 - gt), so we can set the vertical velocity equation equal to zero and solve for t:
0 = 21.0 - 9.81t
Solving for t, we get:
t = 2.14 seconds
Now we can substitute this value of t back into the original equation for h to find the maximum height:
h = 0.8 + (21.0)(2.14) - (1/2)(9.81)(2.14^2)
h ≈ 21.2 meters
Therefore, the maximum height the blue ball reaches is approximately 21.2 meters.
2) To find the height of the blue ball 1.9 seconds after the red ball is thrown, we use the same equation for vertical displacement:
h = h0 + v0t - (1/2)gt^2
In this case, the initial height (h0) is 0.8 meters, the initial velocity (v0) is 21.0 m/s, and the acceleration due to gravity (g) is -9.81 m/s^2.
Plugging in these values and t = 1.9 seconds, we have:
h = 0.8 + (21.0)(1.9) - (1/2)(9.81)(1.9^2)
h ≈ 8.7 meters
Therefore, the height of the blue ball 1.9 seconds after the red ball is thrown is approximately 8.7 meters.
3) To find the time at which the two balls are at the same height, we need to set their respective height equations equal to each other:
h_blue = h_red
Using the equation for vertical displacement, we can write:
h_blue = 0.8 + (21.0)(t) - (1/2)(9.81)(t^2)
h_red = 24.6 + (8.3)(t) - (1/2)(9.81)(t^2)
Setting these two equations equal to each other, we have:
0.8 + (21.0)(t) - (1/2)(9.81)(t^2) = 24.6 + (8.3)(t) - (1/2)(9.81)(t^2)
Simplifying and rearranging the equation, we get:
12.7t = 23.8
Solving for t, we have:
t ≈ 1.87
Therefore, the two balls are in the air at the same height approximately 1.87 seconds after the red ball is thrown.