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phy 121
Your 'cq_1_08.2' 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_08.2_labelMessages **
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: ->->->->->->->->->->->-> :
Vf = V0 +a 'dt
'dt = (Vf-V0)/a
'dt = (0m/s - 15m/s) / -10m/s^2
'dt = 2 s
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The correct solution to the above would be `dt = 1.5 s, which could also have been reasoned directly.
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'ds = (V0+Vf) / 2 * 'dt
'ds = (15m/s + 0) / 2 * 2s
'ds = 15 m
Since it was thrown at 12 m above the ground, and traveled an additional 15 m up, it travels to 27 m above the ground
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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: ->->->->->->->->->->->-> :
'ds = V0'dt + .5a 'dt^2
27m = 0 + .5(10m/s2) * 'dt^2
sqrt(27m/5m/s^2) = 'dt
'dt = 2.32 s
Vf = V0 + a'dt
Vf = 0 + 10m/s^2 (2.32s)
Vf = 23.2 m/s
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
Vf - V0 / a = 'dt
5m/s - 15m/s / -10m/s^2 = 1 s So the ball's speed will be 5m/s at clock time 1 s
However, since it asks about speed not velocity, there is another time to consider: the time at which the ball is at -5m/s velocity or in other words, when the ball is coming down at a speed of 5 m/s
so Vf - V0 / a = 'dt
-5 - 0 / 10m/s^2 = .5 s
So .5 s after the ball starts coming back down, the speed is again at 5 m/s
It reaches max height at t = 2 sec according to a previous solution, so 2s + .5 s = 2.5 s
So the solutions are at clock times 1 s and 2.5 s
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After 1, 2, 3 s the ball's velocity would be 5 m/s, -5 m/s and -15 m/s.
So speed is 5 m/s at t = 1 s and at t = 2 s.
Equations are unnecessary in this case, but would be convenient if the numbers were messier.
In that case you would use say that
speed = | vf | = | v0 + a `dt |
which breaks into the equations
v0 + a `dt = | vf |
and
v0 + a `dt = - | vf |.
In the present case that comes down to
v0 + a `dt = 5 m/s
and
v0 + a `dt = -5 m/s.
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At what clock time(s) will the ball be 20 meters above the ground?
How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
The ball at 20 meters above the ground:
If it starts at 12 m high, it only need to travel 8 m to be a t 20 m high.
So Vf^2 = (15m/s^2) + 2(10m/s^2) * 8m = 8.06 m/s
'dt = 'ds (2/(vf +v0))
'dt = 8m * (2/8.06m/s + 15m/s) = 1.44 s
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You could calculate this relative to the starting point using a single equation.
`ds = v0 `dt + 1/2 a `dt^2
This is a quadratic equation in `dt, which can be put into the standard form of a quadratic as
1/2 a `dt^2 + v0 `dt - `ds = 0.
The quadratic formula yields solutions
`dt = (-v0 +- sqrt( v0^2 - 2 a `ds) ) / (1/2 a).
The advantage to using this equation is that an error in calculating maximum height won't affect the result, which would give you another consistency check.
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For 20 meters high on the way down:
The ball reached max height of 27 m, so it only need to fall 7 m to be at 20 m
Vf^2 = 0 + 20m/s^2 * 7 m (acceleration is positive her becuase the ball is moving in the same direction as the force of gravity)
Vf = 11.83 m/s
'dt = 'ds(2/vf + v0) = 7m * (2/11.83m/s+0m/s) = 1.18 s
After 6 sec:
Vf = V0 + a*'dt
Vf = 15m/s + -10m/s^2 * 6s
Vf = -45m/s
'ds = V0+Vf / 2 * 'dt
'ds = 15-45/2 *6 = -90 m
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Good, but you have an arithmetic error early which affect just about all your subsequent results.
Check my notes, and see also the discussion at the link below.
No need to submit anything more unless you have questions.
&#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. &#
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