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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.
** CQ_1_08.1_labelMessages **
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: ->->->->->->->->->->->-> :
15m/s
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
5m/s
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
Im confused by this question. Does it mean from the moment it was tossed to the end of second 2. or is it just the interval from 1 sec to 2 sec?
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The first two seconds start with the toss and end 2 seconds later.
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I assume from the moment its thrown to the end of 2 sec.
Vave = (25m/s + 5m/s) / 2 = 15 m/s
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You interpreted this correctly.
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
'ds = Vave *'dt
'ds = 15m/s * 2 s = 30 m
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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: ->->->->->->->->->->->-> :
5m/s - (10m/s) = -5m/s So at this point the ball is coming back down at 5m/s
After another second it would be -15 m/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: ->->->->->->->->->->->-> :
'dt = (Vf-V0)/a so 'dt = (0m/s-25m/s) / 10m/s^2 = 2.5s
'ds = Vave * 'dt = (25m/s + 0 m/s) / 2 * 2.5s = 31.25 m
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You could directly reason this.
Losing 10 m/s every second it takes 2.5 seconds to come to rest. During this time average velocity is (25 m/s = 0) / s = 12.5 m/s. 2.5 s * 12.5 m/s = 31.25 m/s.
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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: ->->->->->->->->->->->-> :
Vf = V0 + (a*'dt
Vf = 25m/s + (-10m/s^2 * 4 s)
Vf = -15m/s
So Vave = (Vf+V0) / 2 = (-15m/s + 25m/s) / 2 = 10m/s
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Again direct reasoning would work. You figured out the -15 m/s velocity earlier, and the 5 m/s ave vel follows just as you have calculated it (though you appear to have overlooked the division by 2).
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'ds = Vave * 'dt = 10m/s * 4s = 40m
However, since the peak height is 31.25m, and 'ds in 4 s = 40m, this really means that it traveled 31.25 up, and some other distance down to give 40 total meters traveled in 4 seconds.
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Good. It's very important to check consistency of answers, as you have done.
You overlooked division by 2 earlier, which would have given you ave vel 5 m/s rather than 10 m/s.
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So the height would be 40m - 31.25m = 8.75m above the ground
I am not sure if there is a more formal way of going about finding the height, or if reasoning it out is the best way
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
Vf = 25m/s + (-10m/s^2 * 4s)
Vf = -15m/s
'ds = Vave * 'dt
'ds = -10m/s *6s
'ds = -60 m
I am not sure what the negative displacment means. But if max height was 31.25m, the ball went 31.25 meters up, and then down another 31.25m wouldnt the displacement just be 0? or would it be 31.25 + 31.25 = 62.5m? So does 'ds = -60m mean it fell an additional 60 m from where it was originially thrown from?
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All these calcuations would give you results relative to the initial position.
Either there's a hole for the object to fall down, or the result is meaningless due to the fact that acceleration changes when the ball hits the ground.
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15-20
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Please see my question on the last problem
&#Good responses. See my notes and let me know if you have questions. &#