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?
• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
• At what clock time(s) will the speed of the ball be 5 meters / second?
• 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: ->->->->->->->->->->->-> :
By setting the final velocity equal to 0, the time required to make it to the highest point will be 0=(-10*dt)+15m/s. The time required equals 1.5 seconds. The average velocity will be equal to 7.5m/s at 1.5 seconds which will produce a displacement of 11.25m+12m which equals 23.25m.
By solving for the final velocity using a distance of -23.25, I come up with a final velocity of -26.3m/s. the average velocity is equal to -5.65m/s. Therefore the time required is 4.12 seconds.
By setting vf = to 5m/s, I come up with a clock time of 1 second.
The ball will be 20 meters above the ground at 1.1s because it starts at 12m above ground..
At the end of 6 seconds, the ball will be on the ground.
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20 min.
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I was having trouble with the last couple questions
I believe you trouble was mainly with interpretation. No need for a revision but check the discussion at the link below:
&#Please compare your 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. &#