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PHY 231
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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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: ->->->->->->->->->->->-> :
0=(15 m/s)^2+2*-10 m/s^2 *^' ds
11.25m='ds
11.25m=(0+15 m/s)/2 *^' dt
1.5='dt
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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: ->->->->->->->->->->->-> :
vf^2=0+2*-10 m/s^2 *(12m+11.25m)
vf=21.56 m/s
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
5 m/s=15 m/s+-10 m/s^2 *^' dt
1 s=^' dt
5 m/s=0 m/s+-10 m/s^2 *^' dt
'dt=0.5 s+1s=1.5s
@& The speed is also 5 m/s when the velocity is -5 m/s, which with your sign convention occurs on the way down.*@
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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: ->->->->->->->->->->->-> :
0 m/s=15 m/s+(-10m/s^2 ) *^' dt
0.67 s
And
21.56 m/s=0 m/s+(-10m/s^2 ) *^' dt
2.156 s
'ds=0m/s*6s+.5*-10 m/s^2 *(6s)^2=180 m
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15 min
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&#Good responses. See my notes and let me know if you have questions. &#