Phy 121
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.
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:
Y = v^2 - V0^2 / 2a = 0 - 15m/s^2 / 2*10m/s^2 = 11.3m
T = 15m/s / 5m/s^2 = 3s
change in velocity could be 15 m/s, but the acceleration would be 10 m/s^2, not 5 m/s^2.
V = v0 + at = 15m/s - 10m/s * 3s = -15m/s
T = 15m/s / 5m/s^2 = 3s
your 3 s time interval was your solution for the time to reach the highest point, not the time to reach the ground
T = 1s
This is one possible answer; there is another
T =
Distance at end of 6 sec is 11.3 *2 = 22.6m
this isn't correct
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40min
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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. &#