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.
They are all coverd in water 24/7 they never clear up
The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.
<h3>What is velocity?</h3>
Velocity is a vector quantity that tells the distance an object has traveled over a period of time.
Displacement is a vector quality showing total length of an area traveled by a particular object.
Imagine a time-position graph where the velocity of an object is constant. What will be observed on the graph concerning the slope of the line segment as well as the velocity of the object?
The slope of the line is equal to zero and the object will be stationary.
The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.
To learn more about velocity refer to the link
brainly.com/question/18084516
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