#$&*
PHY 121
Your 'cq_1_08.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.
What will be the velocity of the ball after one second?
answer/question/discussion: ->->->->->->->->->->->-> n:
v0 = 25m/sec
a= 10m/sec^2
`dt = 1sec
Vf = v0 + (a*`dt)
Vf = 25m/sec + (10m/sec^2 * 1sec)
Vf = 25m/sec +10m/sec
Vf =35m/sec
@&
The ball is going upward, the acceleration is downward. The ball will be slowing down.
*@
#$&*
What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> n:
Vf = v0 + (a*`dt)
Vf = 25m/sec + (10m/sec^2 * 2sec)
Vf = 25m/sec + 20m/sec
Vf = 45m/sec
#$&*
During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> n:
vAve = (25m/sec +45/sec)/2
vAve = 35m/sec
#$&*
How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> n:
`ds = vAve * `dt
`ds = 35m/sec * 2sec
=70m
#$&*
What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> n:
Vf = v0 + (a*`dt)
Vf = 25m/sec + (10m/sec^2 * 3sec)
Vf = 25m/sec + 30m/sec
Vf = 55m/sec
Vf = v0 + (a*`dt)
Vf = 25m/sec + (10m/sec^2 * 4sec)
Vf = 25m/sec + 40m/sec
Vf = 65m/sec
#$&*
At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> n:
vf = v0 + a * `dt
65m/sec = 25m/sec + (10m/sec^2 * `dt)
40m/sec = 10m/sec^2 * `dt
(40m/sec)/10m/sec^2 = `dt
`dt = 4sec
`ds = (v0 + vf)/2 * `dt
`ds = (25m/sec + 65m/sec)/2 * 4sec
`ds = (90m/sec)/2 * 4sec
`ds = 180m
#$&*
What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
answer/question/discussion: ->->->->->->->->->->->-> n:
aAve = `dv/`dt
=(65m/sec - 25m/sec)/4sec
=10m/sec^2
`ds = (25m/sec + 65m/sec)/2 * 4sec
`ds = 180m
#$&*
How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> n:
`ds = (v0 * `dt) + .5(a*`dt^2)
`ds = (25m/sec * 6sec) + .5(10m/sec^2 * (6sec)^2)
=150m+ .5(60m)
=180m
#$&*
*#&!
@&
All your reasoning is good but you didn't correctly consider the relative directions of the initial velocity and the acceleration.
No need for a revision, since I'm confident you'll understand the discussion at the link.
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new 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.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#
*@