PHY 201
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A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.
• What will be the velocity of the ball after one second?
answer/question/discussion: ->->->->->->->->->->->-> :
As the ball is going up the velocity is decreasing because of the acceleration of gravity, so the time it takes to reach the top is the exact time it takes to reach the bottom from the top. The velocity can be subtracting the initial velocity from the acceleration of gravity and then multiplied by the time, so: v = v0 + ag * time = 25 m/s – 10 m/s^2 * 1 s = 15 m/s. Since the acceleration of gravity is going downward, then it is a negative.
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
I will use the same method as before: v = v0 + ag * time = 25 m/s – 10 m/s^2 * 2 s = 5 m/s
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
First, I got the average velocity of the first second, which is: (15 m/s + 25 m/s) / 2 = 5 m/s vAve, Then I got the average velocity of the second second, which is (5 m/s + 25 m/s) / 2 = 15 m/s vAve. Finally, I took the two average velocities and took the average of them which is: (5 m/s + 15 m/s) / 2 = 17.5 m/s
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
If you have the both the initial and final velocities, the acceleration, and time interval, then you can find the rest of the variables. First I am going to find the average velocity, which is: (15 m/s + 25m/s) / 2 = 20 m/s. Then, I am going to multiply the average velocity by the time interval and get my displacement: 20 m/s * 1 s = 20 meters. That was for the first second.
This one is for the second second: I used the same method, except my initial velocity, which is different: (5 m/s + 25 m/s) / 2 = 15 m/s, 15 m/s * 2 s = 30 meters.
So, the ball traveled for 50 meters upward.
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• What will be its velocity at the end of an additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> :
3 Seconds: I will use the same method as before: v = v0 + ag * time = 25 m/s – 10 m/s^2 * 3s = -5 m/s
4 Seconds: I will use the same method as before: v = v0 + ag * time = 25 m/s – 10 m/s^2 * 4s = -15 m/s
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• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> :
2.5 seconds, because when you put 2.5 s into the equation it is equal to zero: v = v0 + ag * time = 25 m/s – 10 m/s^2 * 2.5s = 0 m/s
It has risen 31.25 meters in 2.5 seconds to reach to top, because if the average velocity is: (25 m/s + 0 m/s) / 2 = 12.5 m/s, and 12.5 m/s * 2.5 s = 31.25 meters.
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• What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
answer/question/discussion: ->->->->->->->->->->->-> :
The average velocity of the first four seconds is 5 m/s: (25 m/s + -15 m/s) / 2 = 5 m/s.
If the average velocity is 5 m/s and the time interval is 4 seconds, then the change in position is: 5 m/s * 4 s = 20 m.
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
The ball will not be in the air for 6 seconds, because the full time it takes the ball to go up and down is 5 seconds, so there is no 6 seconds.
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1 hour
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Good work. The link below is provided for reference.
&#See any notes I might have inserted into your document. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your solutions with the expanded discussion at the link
Solution
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