#$&*
Phy 201
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
• How high does it rise and how long does it take to get to its highest point?
answer/question/discussion: ->->->->->->->->->->->-> :
v=v0+a*t
0=15m/s-10m/s^2*t
-15m/s=-10m/s^2*t
1.55s=t
t=1.55s= time to reach maximum
vAve= v0+v1.55 / 2 = 15m/s + 0 m/s / 2 = 7.5m/s
7.5m/s*1.5s = 11.25m max height from height thrown
#$&*
• 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: ->->->->->->->->->->->-> :
To see how long the ball took to hit the ground, I first looked at how long it took to reach its maximum height which was 1.5sec. So by drawing a picture I saw that it took 1.5 sec to reach the maximum point, then it took another 1.5s to fall 11.25 m. Then it took a total of two more 1.5s intervals to fall a total of 22.5 m. (1.5s* 4 intervals = 6s). However the total distance from the max height to the ground is 23.25m. (12m+11.25m= 23.25m). Then I computed the total amount of time to travel by doing this simple proportion equation:
22.5m/ 6s = 23.25m/ x
22.5m * x = 139.5 m*s
x= 6.25s = total time after initial toss to hit the ground.
@& Simple proportion won't work. This would only work if the velocity was unchanging. The ball travels that last 12 meters much faster than the first 11.25 m.
You would get the correct answer by using the equations of uniformly accelerated motion..*@
Then to find the speed that the ball was traveling when it hi the ground all I have to do is divide the total distance the ball traveled (23.25m + 11.25m = 34.5m) by the total time it traveled, 6.25s.
34.5m/ 6.25s = 5.52m/s
#$&*
• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
v=v0+a*t
5m/s= 15m/s-10m/s^2 *t
-10m/s=-10m/s^2 * t
t= 1s
#$&*
• At what clock time(s) will the ball be 20 meters above the ground?
• answer/question/discussion:
•
• by using the final velocity I can figure out these clock times:
• 5.52m/s = 20m/x
• x= 3.62s
@& This is not a direct proportionality. The ball's velocity is changing. You need to use the equations of motion.*@
•
• Also since the ball was thrown from a height of 12 m then the ball would have also been 20m from the ground at the distance of 8m so:
•
• 5.52m/s=8m/x
• x=1.45s
•
• #$&*
•
• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
As I calculated earlier by adding up four 1.5sec intervals, at 6sec the ball will be dropped by 22.5m which means that it is 0.75 m from the ground. (23.25m-22.5m= 0.75m)
#$&*
** **
45min
** **
@& You are using direct proportionalities in some situations where they don't apply.
*@
&#See any notes I might have inserted into your document, and before looking at the link below see if you can modify your solutions. If there are no notes, this does not mean that your solution is completely correct.
Then please compare your old and new 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.If your solution is completely consistent with the given solution, you need do nothing further with this problem. &#