Phy 201
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.
** **
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?
* It would take approximately 1.5 seconds for the ball to travel 27 meters up (from the ground) and reach its highest point. I got this first by dividing the v0 of 15m/s by the a of 10m/s to get 1.5 seconds to reach the max height.
Good, but I don't think it reaches quite as high as 27 meters. You didn't explain how you got this result.
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
* Using the formula vf^2 = v0^2 + 2a*'dt, I got the vf on the downfall of this ball to be 23.2m/s. And this would take approximately 3.82 seconds to hit the ground after it was originally tossed. I got this with the equation 'dt = 'dv/a, and then adding the 1.5seconds it took to reach max height.
At what clock time(s) will the speed of the ball be 5 meters / second?
* 1 second.
At what clock time(s) will the ball be 20 meters above the ground?
* At 1.1 seconds, the ball will be 8 meters above the person's hand, and 20 meters above the ground. I got this be using the formula 'ds = vAve * 'dt. I got the vAve to be 7.5m/s by (25m/s + 0m/s)/2 = 7.5m/s. Then plugging that into 8meters = 7.5m/s * 'dt, I get the 'dt to equal 1.1 sec.
The ball will pass the 20 m position twice, once on the way up and once on the way down.
* How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> scussion: Well, I thought I had already established that the ball would only be in the air 3.82 seconds. But if you want me to go into negatives.... the ball would be at 25.2 meters UNDER GROUND. I think...?
You always want to consider positive and negative results. Sometimes one or more results are then rejected as not being consistent with reality.
** **
About 25 minutes...
** **
You do appear to have missed a couple of details, though your overall conception of the problem is very good.
&#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. &#