mth173
Your 'cq_1_8.2' 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?
Ds = vAve * Dt
Ds = (15m/s + 0m/s) / 2 * 1.5s
Ds = 11.25m
15m/s / 10ms/s/s = 1.5s = Dt
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?
t = 1sec
At what clock time(s) will the ball be 20 meters above the ground?
20m / 7.5m/s = 2.7sec
This would be the case of 20 m was max height; but it's not so. Max height occurs after 1.5 sec, at a distance of 22.5 m above the initial point.
How high will it be at the end of the sixth second?
answer/question/discussion: 0m
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15mins
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&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#