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course phy 201
I was really confused on this part. I couldn't find any notes on this subject or any related problems to help.
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:
vf = 25 - 10 x 1
vf = 15m/s
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What will be its velocity at the end of two seconds?
answer/question/discussion:
vf = 25 -10 x 2
vf = 5m/s
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion:
30/ 2 = 15m/s
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Except for the lack of units, you're good so far.
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How far does it therefore rise in the first two seconds?
answer/question/discussion:
-20 / 2 = -10
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If you think about it you know the initial and final velocities for the first two seconds. How can you use that information to find the displacement?
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What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion:
vf = 25 - 10 x 3
vf = -5m/s
vf = 25 - 10 x 4
vf = -15m/s
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At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion:
‘ds = 25 x 1 + .5 x -9.8 x 1^2
‘ds = 20.1m
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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:
aAve = 10 /2 = 5m/s
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You mean vAve.
You don't say how you got the 10 to divide by 2. You also haven't included units. But 10 m/s would be the correct numerator, so I expect that you did the right thing to get that.
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5 x 4 = 20m
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5 m/s * 4 s = 20 m.
5 * 4 = 20 and nothing in that expression has any units or any meaning.
That said, except for the units you did well on this question.
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How high will it be at the end of the sixth second?
answer/question/discussion:
aAve = -10 / 2 = -5m/s
-5 x 6 = -30m
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A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
How high does it rise and how long does it take to get to its highest point? answer/question/discussion:
12 = .5 x 9.8 x ‘dt^2
12 = 4.9dt^2
dt = 1.5s
15 x 1.5 = 22.5m + 12 = 34.5m
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15 * 1.5 = 22.5, not 34.5.
22.5 + 12 = 34.5, but 15 * 1.5 doesn't.
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How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground? answer/question/discussion:
34.5m x 10m/s^2 = 345m/s
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Nothing is moving at 345 m/s.
34.5 m * 10 m/s^2 = 345 m^2 / s^2, but this is not relevant. Position * acceleration has no important meaning.
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34.5 / 10 = 3.45s
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At what clock time(s) will the speed of the ball be 5 meters / second? answer/question/discussion:
2s
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right
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At what clock time(s) will the ball be 20 meters above the ground?
.5s
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at t = .5 s the ball will be moving at 20 m/s, its average velocity will have been 22.5 m/s and it will have rise 11.25 meters. It iwll be close to 20 m off the ground. but not exactly.
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How high will it be at the end of the sixth second?answer/question/discussion:
On the ground
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