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Phy 201
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cq_1_082
#$&*
phy 201
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Seed 8 2
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
• How high does it rise and how long does it take to get to its highest point?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
t0=12 m
a= -10m/s/s
v0=15 m/s
vf=0 m/s
• vf = v0 + a * `dt
vf-v0)/a=dt
-15/-10=dt
Dt=1.5s
vAve=7.5 m/s *1.5 s
it rises 11.25 m to a maximum height of 26.25 meters and it takes 1.5 seconds to get there
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Good. It does rise 11.25 m. But having started from the 12 m position it rises only to a 23.25 m height.
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That was a careless mistake I saw the the 15 at the top and used it rather than 12.
• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
Ds=-26.25 m
ds = (v0 + vf) / 2 * `dt
vf = v0 + a * `dt
vf= 15 m/s -10m/s/s *4 s
vf= -25 m/s
ds-vAve=dt
?????
I don’t know how to do this
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
#$&*
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Good try.
The ball started from the 12 m position so its displacement, from that position, is only -12 m. You also don't know that it takes 4 seconds to get there.
What you do know, therefore, is that
`ds = -12 m
v0 = +15 m/s
a = -10 m/s^2.
7&&&&&&&&&&&
vf^2 = v0^2 + 2 a `ds
vf^2= (15 m/s^2)+ 2(-10m/s/s)(-12m)
sqrt(465)
vf=21.56m/s when it hits the ground
• vf = v0 + a * `dt
vf-v0=a*dt
vf-v0 / a= dt
dt=21.56- 15=6.56/-10=
.656 seconds
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This would be -.656 seconds, not .656 seconds.
The 21.56 m/s final velocity is in the direction opposite the 15 m/s initial velocity, so the two have opposite signs.
Since you're using -10 m/s^2 for acceleration, it's clear that you've chosen upward as your positive direction, so vf = -21.56 m/s.
That changes your calculation to
`dt = (-21.56 m/s - 15 m/s) / (-10 m/s^2) = -3.156 sec.
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Im still not very confident in my revisions.
Now which equation(s) contain these three variables, and which will be the easiest to solve for the remaining variable?
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
5m/s = 15 m/s-10 m/s^2 *dt
5m/s-15m/s)/-10=dt
At 1 second the speed will be at 5m/s
#$&*
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That's true. So far so good.
But there's another time at which the speed is 5 m/s. Hint: When the velocity is -5 m/s, the speed is 5 m/s.
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• At what clock time(s) will the ball be 20 meters above the ground?
ds = (v0 + vf) / 2 * `dt
8 - 7.5)*2= 1 second
• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> : ->->->->->->->->->->->-> :
• `ds = v0 `dt + .5 a `dt^2
Ds= 15 m/s (6) +.5(-10m/s^2)(6^2)
90 + -180
-90 m is how high it would be. Im assuming it would’ve already come down and it the ground,
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Good attempts. As you know you didn't get everything, especially on the first problem. I don't think you'll have much trouble going a little further.
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&#Good responses. See my notes and let me know if you have questions. &#