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PHy121
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: ->->->->->->->->->->->-> :
IT will rise for 1.5 seconds (15/10). It will reach a height of 34.5meters (15m/sec *1.5 sec + 12m starting point).
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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: ->->->->->->->->->->->-> :
It takes four seconds to reach the ground (three for the full cycle from starting point and another second from the starting point to the ground).
I do not understand how to calculate speed here? I'm not even 100% sure that my logic is correct in anything I've written above.
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
15m/sec - 10m/sec = 5m/sec It will occur after one second.
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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: ->->->->->->->->->->->-> :
at about 0.5 seconds, the ball will be 20 meters above ground. It starts at 12m and has a v0 of 15m/sec. The 8 meters difference would take about one half second.
The object will be on the ground at the end of the sixth second.
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c082
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