Your 'cq_1_8.1' 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?
vf^2 = v0^2 + 2 a `ds.
a=-10
'ds=1
Vo=25
vf^2=625 +-20
Vf=24.6m/s
`dt = 1 second.
`ds is not equal to 1, and `ds has units. Had you used units you probably would have detected the error.
You don't need equations to answer this question. It's OK to use an equation, but you should also be able to reason this out by common sense. The arithmetic can easily be done in your head.
What will be its velocity at the end of two seconds?
vf^2 = v0^2 + 2 a `ds.
a=-10
'ds=2
Vo=25
vf^2=625 +-40
Vf=24.2m/s
During the first two seconds, what therefore is its average velocity?
(21.9+21.4)/2=21.65 m/s=Vave
How far does it therefore rise in the first two seconds?
`ds = v0 `dt + .5 a `dt^2
'ds=25 x 2 + .5 -10 4
'ds=30m
Note that this is inconsistent with your assumption `ds = 2.
What will be its velocity at the end of a additional second, and at the end of one more additional second?
vf^2 = v0^2 + 2 a `ds.
a=-10
'ds=3
Vo=25
vf^2=625 +-60
Vf=23.8m/s
vf^2 = v0^2 + 2 a `ds.
a=-10
'ds=4
Vo=25
vf^2=625 +-80
Vf=23.3m/s
At what instant does the ball reach its maximum height, and how high has it risen by that instant?
vf^2 = v0^2 + 2 a `ds.
a=-10
'ds=2
Vo=25
Vf=0
vf^2=500 +-40
0=25^2 + 2 -10 'ds
'ds= 31.25 seconds
`ds = v0 `dt + .5 a `dt^2
'ds=25 31.25 + .5 -10 25^2
'ds=3906.3 m
What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
Vave=24.15
20m
How high will it be at the end of the sixth second?
'ds=25x 6 +.5 -10 36
'ds=30m
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15min
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You misidentified at least one of your variables. You should see my notes and give the problem a few minutes' thought before seeing the solution below:
The ball starts out at 25 m/s and changes velocity by -10 m/s every second. So after 1 second it is traveling at 15 m/s, and after 2 seconds at 5 m/s.
Its average velocity for the first second is the average of its 25 m/s initial velocity and its 15 m/s final velocity on that interval. So average velocity is 20 m/s and the displacement is `ds = vAve `dt = 20 m/s * 1 s = 20 m.
Its average velocity for the first 2 seconds is the average 15 m/s of its initial 25 m/s velocity and its final 5 m/s velocity on this interval, and its displacement is `ds = vAve * `dt = 15 m/s * 2 = 30 m.
After 4 seconds it is moving at -15 m/s and its average velocity for the 4-second interval is 5 m/s, the average of 25 m/s and -15 m/s. Its displacement is 5 m/s * 4 s = 20 m.
It reaches its maximum height at the instant it comes to rest. To come to rest from 25 m/s, accelerating at -10 m/s^2, requires 2.5 seconds. The average velocity on this interval is 12.5 m/s, the average of its 25 m/s initial velocity and its 0 m/s final velocity on this interval. At an average of 12.5 m/s the ball will travel 12.5 m/s * 2.5 s = 31.25 meters, and this is the maximum height to which it will rise.