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?
After 1 second the ball will decrease its speed by 10m/sec^2 from the initial 25m/sec. So the velocity will be 15 m/ sec
answer/question/discussion:
• What will be its velocity at the end of two seconds?
After 2 seconds the ball will turn around. So the velocity will be at 5 m/ sec
The ball won't have turned around by this time; velocity will be 5 m/s
answer/question/discussion:
• During the first two seconds, what therefore is its average velocity?
The velocity at t= 2 seconds is -5 m /sec. The initial is 25 m /sec
VAve= (v0+vf)/2
vAve = (25 + 5)/2
vAve = 30/2
vAve = 15 m /sec
answer/question/discussion:
• How far does it therefore rise in the first two seconds?
‘ds = vAve * ‘dt
‘ds = 15m/sec * 2 seconds
‘ds = 30m
answer/question/discussion:
• What will be its velocity at the end of a additional second, and at the end of one more additional second?
To get the velocity at t=3 seconds, the ball will reach 0m/sec and change direction. The acceleration is still 10m/sec^2. At t= 3, the velocity will be -5 m /sec. At t=4, the velocity of the ball will be -15 m /sec
answer/question/discussion:
• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
The ball will reach its maximum height at about 2.5 seconds. After 2 seconds the ball is traveling at 5m/sec and after another half second it reaches 0/sec, and therefore its maximum height. To calculate how high the ball goes, I first calculate the Average velocity. The final velocity is 0m/sec, so the average is 12.5 m/sec over a period of 2.5 seconds. I then use the equation ‘ds = vAve * ‘dt
‘ds = 12.5m/sec * 2.5 seconds
‘ds = 31.25 m
answer/question/discussion:
• What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
At t=4, v = -15 m /sec.
vAve = (v0+vf)/2
vAve = (25-15)/2
vAve = 10m/sec
(25-15)/2 = 10 / 2 = 5
This means that at this point the ball has gone up more than it has gone down. The ball traveling up at a constant rate of 10m/sec would be an equivalent displacement for what actually happened(the ball went up then down.)
‘ds = vAve * ‘dt
‘ds = 10m/sec * 4sec
‘ds = 40m
answer/question/discussion:
• How high will it be at the end of the sixth second?
At t= 6, v = -35 m /sec
vAve=(v0+vf)/2
vAve=(25 -35)/2
vAve=-10/2
vAve= -5m/sec
‘ds = vAve * ‘dt
‘ds = -5m/sec * 6sec
‘ds = -30 m
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20 minutes
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I cannot explain my second to last answer of 40m. 40m is higher than the highest point the ball went before it went down. What am I missing?
Just an arithmetic error. See my note above.
&#Good responses. See my notes and let me know if you have questions. &#