#$&*
Phy 121
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_08.2_labelMessages **
Copy the problem below into a text editor or word processor.
• This form accepts only text so a text editor such as Notepad is fine.
• You might prefer for your own reasons to use a word processor (for example the formatting features might help you organize your answer and explanations), but note that formatting will be lost when you submit your work through the form.
• If you use a word processor avoid using special characters or symbols, which would require more of your time to create and will not be represented correctly by the form.
• As you will see within the first few assignments, there is an easily-learned keyboard-based shorthand that doesn't look quite as pretty as word-processor symbols, but which gets the job done much more efficiently.
You should enter your answers using the text editor or word processor. You will then copy-and-paste it into the box below, and submit.
________________________________________
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
• How high does it rise and how long does it take to get to its highest point?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = v0 + a * ‘dt
vf = 15m/s + -10m/s^2 * 1.5s
vf = 15m/s -15m/s
vf = 0m/s
vAve = (v0 +vf) / 2
vAve = (15m/s + 0m/s) / 2
vAve = 15m/s / 2
vAve = 7.5m/s
The ball reaches its highest point at 1.5seconds with an average velocity of 7.5m/s, this ball will have traveled 11.25m because 7.5m/s * 1.5s = 11.25m.
#$&*
• 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: ->->->->->->->->->->->-> :
vf^2 = v0^2 + 2 * a * ‘ds
vf^2 = (15m/s)^2 + 2 * -10m/s^2 * 34.5m
vf^2 = 225m^2/s^2 + -690m^2/s^2
vf^2 = -465m^2/S62
‘sqrt(vf^2) = ‘sqrt(-465m^2/s^2)
vf = +-21.6m/s
In this case it would be a negative velocity.
vf = v0 + a * ‘dt
-21.6m/s = 15m/s + -10m/s^2 * ‘dt
-36.6m/s = -10m/s^2 * ‘dt
‘dt = 3.66 seconds.
When the ball hits the ground its velocity is -21.6m/s and from the initial toss it will take 3.66 seconds for the ball to reach the ground.
#$&*
• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = v0 + a * ‘dt
5m/s = 15m/s + -10m/s^2 * ‘dt
-10m/s = -10m/s^2 * ‘dt
1s = ‘dt
-5m/s = 15m/s + -10m/s^2 * ‘dt
-20m/s = -10m/s^2 * ‘dt
2s = ‘dt
When the ball has a speed of 5m/s the clock times will be 1s and 2s.
#$&*
• 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: ->->->->->->->->->->->-> :
vf^2 = v0^2 + 2 * a * ‘ds
vf^2 = (15m/s)^2 + 2 * -10m/s^2 * 8m
vf^2 = 225m^2/s^2 + -160m^2/s^2
vf^2 = 65m^2/s^2
‘sqrt (vf^2) = ‘sqrt (65m^2/s^2)
vf = +-8.1m/s
vf = v0 + a * ‘dt
8.1m/s = 15m/s + -10m/s^2 * ‘dt
-6.9m/s = -10m/s^2 * ‘dt
.69s = ‘dt
vf = v0 + a * ‘dt
-8.1m/s = 15m/s + -10m/s^2 * ‘dt
-23.1m/s = -10m/s^2 *’dt
2.31s = ‘dt
The clock times for when the ball is 20 meters from the ground are .69s and 2.31seconds.
-vf = v0 + a * ‘dt
vf = 15m/s + -10m/s^2 * 6s
vf = 15m/s + -60m/s
vf = -45m/s
‘ds = (v0 + vf) / 2 * ‘dt
‘ds = (15m/s + -45m/s) / 2 * 6s
‘ds = -30m/s / 2 * 6s
‘ds = -15m/s * 6s
‘ds = -90m
At the sixth second the ball will be -90 meters.
#$&*
*#&!
&#Good responses. Let me know if you have questions. &#