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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.
** CQ_1_08.2_labelMessages **
Asst_8 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: ->->->->->->->->->->->-> sion:
When vf=0m/s the ball will have reached its max height
Known variables are therefore:
v0= 15m/s
a=-10m/s^2
vf= 0m/s
using vf=v0 + a*dt and solving for dt:
0m/s = 15m/s + -10m/s^2 * dt
-15m/s = -10m/s^2 *dt
-15m/s / -10m/s^2 = dt
1.5s = dt
It takes 1.5 seconds for the ball to reach its max height.
To calculate the total height the ball rises:
`ds = v0`dt + .5a*`dt^2
ds= 15m/s(1.5s) + .5(-10m/s^2)*(1.5s)^2
ds= 22.5m + -5m/s^2 * 2.25s^2
ds= 22.5m + -11.25m
ds= 11.25m
To calculate total height I add the 12 meters that it initially was, so 11.25m +12 m = 23.25m
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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: ->->->->->->->->->->->-> sion:
ground level is presumeably 0m. After 1.5 seconds, the ball is farthest from the ground at 23.25m.
v0=15m/s
ds= 23.25m + 11.25m = 34.5m
a=-10m/s^2
to find vf, I used vf^2 = v0^2+2a`ds:
vf^2 = (15m/s)^2 + 2(-10m/s^2)*34.5m
vf^2 = 225(m/s)^2 + -20m/s^2 * 34.5m
vf^2 = 225(m/s)^2 - 690(m/s)^2
vf^2 = - 465(m/s)2
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Good, except that the equation required the displacement of the object, not the distance it travels.
The displacement is the change in position from the first instant to the last, not the total distance traveled.
To solve directly from the given information note that from the initial to the final point the displacement was 12 m downward. Letting upward be the positive direction, then, `ds was -12 m.
v0 was +15 m/s and a was -10 m/s^2.
Following your reasoning with these quantities would lead you to the correct final velocity. Your right-hand side, using these quantities, will end up positive so it will be possible for you to take the square root.
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> sion:
I figured this based on vf=5m/s and solved for `dt to figure at what clock time the ball reaches this speed
v0=15m/s
vf=5m/s
a=-10m/s^2
using vf=v0 + a*dt and solving for dt:
5m/s = 15m/s + -10m/s^2 * dt
-10m/s = -10m/s^2 *dt
-10m/s / -10m/s^2 = dt
1s = dt
After one second, the speed of the ball will be 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: ->->->->->->->->->->->-> sion:
To find clocktime when ball is 20m above ground I first subtracted the 12m for the initial height of the ball to find a `ds of 8m:
Next I calculated solved for vf to find how fast the ball traveled at the 8m height using the following formula:
vf^2= v0^2 + 2a*`ds
vf^2= 225(m/s)^2 + -20m/s^2* 8m
vf^2= 225(m/s)^2 -160(m/s)^2
vf^2= 65(m/s)^2
vf=+-sqrt(65(m/s)^2)
vf(apx)=+-8m/s
I then plugged this value for vf into vf=v0+a*`dt and solved for `dt:
8m/s = 15 m/s + -10m/s^2*`dt
-7m/s = -10m/s^2 * `dt
-7m/s / -10m/s^2 = `dt
7/10s = `dt
`dt(apx)= .7s
The ball was 20meters above ground at .7 seconds.
To find the displacement of the ball after 6 seconds:
`ds= 15m/s*6s + .5(-10m/s^2)*(6s)^2
`ds= 90m + -5m/s^2(36s^2)
`ds= 90m + - 180m
`ds= 90m
`ds is 90m however I add the initial height of 12m to calculate after 6seconds the ball is -78m below the surface of the ground (theoretically or if there was a canyon)
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45minutes
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Overall good, but the distinction between distance and displacement caused you at least one problem. Should be easy to correct.
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