phy 201
Your 'cq_1_08.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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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: ->->->->->->->->->->->-> :
acceleration = final velocity - initial velocity / time
final velocity = acceleration * time + initial velocity
= 10 m/s^s * 1sec + 25m/s
= 35m/s
What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
acceleration = final velocity - initial velocity / time
final velocity = acceleration * time + initial velocity
= 10 m/s^s * 2sec + 25m/s
= 45m/s
During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
average velocity = 35 + 45/2 = 40m/s
How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
distance = average velocity * time = 40 * 2 = 80 meters
What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> :
acceleration = final velocity - initial velocity / time
final velocity = acceleration * time + initial velocity
= 10 m/s^s * 3sec + 25m/s
= 55m/s
acceleration = final velocity - initial velocity / time
final velocity = acceleration * time + initial velocity
= 10 m/s^s * 4sec + 25m/s
= 65m/s
At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> :
acceleration = finial vleocity - initial velocity /time
time = 65m/s - 25m/s / 10m/s^2
= 4 sec
distance = average vleocity * time
= 35+45+55+65/4 * 4 sec
= 200 meter
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: ->->->->->->->->->->->-> :
average velocity = 35+45+55+65/4 = 50meter
distance = average vleocity * time
= 35+45+55+65/4 * 4 sec
= 200 meter
How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
distance = average vleocity * time
= 35+45+55+65/4 * 6 sec
= 300 meter
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15 minutes
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All your work is very good, assuming acceleration to be positive.
However the initial velocity is upward and the acceleration is downward. So one will be positive and the other negative. Which is which depends on your choice of the positive direction.
No need to resubmit, provided you understand everything at the link below:
&#Please compare your solutions with the expanded discussion at the link
Solution
Self-critique your solutions, if this is necessary, according to the usual criteria. Insert any revisions, questions, etc. into a copy of this posted document. Mark any insertions with &&&& so they can be easily identified. &#