It is often revealed <span>at the resolution of the story, when the reader can see how the story ends.</span>
Power = work / time = 8000J / 20s = 400W
I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:
When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.
1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²
For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.
2. For this, you equate the y values of both balls:
y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t,
t = 2.25 seconds
Thus, the two balls would be at the same height after 2.25 seconds.
Apparent magnitude depends mainly on the brightness of the object as seen from an observer on Earth. This is taken into account without the effects of the atmosphere.
