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phy 121
Your 'cq_1_07.1' 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_07.1_labelMessages **
A ball falls freely from rest at a height of 2 meters. Observations indicate that the ball reaches the ground in .64 seconds.
• Based on this information what is its acceleration?
answer/question/discussion: ->->->->->->->->->->->-> :
v_0=0m/s
'dt=0.64 sec
'ds=2 meters
vAve='ds/'dt
vAve=(2 m)/(0.64 sec)
=3.125m/s
vAve*2=v_f
v_f=3.125m/s*2
v_f=6.25m/s
Now I can find 'dv=(v_f-v_0)
'dv=(6.25m/s-0m/s)
'dv=6.25 m/s
aAve='dv/'dt
=9.77 m/s^2
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• Is this consistent with an observation which concludes that a ball dropped from a height of 5 meters reaches the ground in 1.05 seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
NO. Ball dropped at v_0= m/s, 'ds=5 m, 'dt=1.05 sec then the acceleration is
vAve='ds/'dt
vAve=(5m)/(1.05 sec)
=4.76m/s
vAve*2=v_f
v_f=4.76m/s*2
v_f=9.52m/s
aAve='dv/'dt
'dv=(v_f-v_0)
'dv=9.52 m/s
aAve=(9.52m/s)/(1.05 sec)
aAve=9.07 m/s^2
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• Are these observations consistent with the accepted value of the acceleration of gravity, which is 9.8 m / s^2?
answer/question/discussion: ->->->->->->->->->->->-> :
The 9.77 m/s^2 from the first observation is consistent with the accepted value of acceleration of gravity of 9.8 m/s^2.
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20 min
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&#Good responses. Let me know if you have questions. &#