Phy 121
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.
** **
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: ->->->->->->->->->->->-> :
To determine the velocity of the ball after 1 second I used the equation y=v0t+1/2at^2
When plugging in my numbers this gave me 25m/s(1.0s)+1/2(10m/s^2)
Which becomes 25m/s+10m/s^2 (1sec)= 35m/s
The ball's velocity is upward and acceleration is downward.
All quantities must be expressed in the context of the chosen positive direction, so the first thing you needed to do here was choose your positive direction. Upward or downward is fine, but you have to choose one.
Then your initial velocity and acceleration, being in opposite directions, will have opposite signs.
#$&*
• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
At the end of two seconds the velocity would be 45m/s
25m/s+(10m/s^2)(2sec)=45 m/s
#$&*
• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
The average velocity is 10 m/s
Vf-v0/2= average velocity
45m/s-25m/s= 20 m/s/2 = 10m/s
When you average two numbers you don't subtract.
When you find the change between two quantities you do subtract.
#$&*
• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
In two seconds it will rise 20 m/s.
This is the change in velocity during the first two seconds.
#$&*
• What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> :
45m/s+(10m/s^2)(1sec)= 55m/s After the end of an additional second.
55m/s+ (10m/s^2)(1sec)= 65m/s at the end of one more additional second.
#$&*
• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> :
The ball reaches its maximum height at 4 seconds. Over the course of that time the ball has risen 50 meters.
#$&*
• 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 4 seconds is as follows:
65m/s-25m/s/2 = 40m/s/2= 20 m/s as the average velocity
#$&*
• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
60 meters by the end of the 6th second.
#$&*
** **
30 minutes
** **
You're doing a good job of applying rules to the analysis.
Sometimes you get the rules wrong, or neglect important details (e.g., choosing a positive direction and giving quantities signs consistent with the chosen positive direction). See my notes.
&#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. &#