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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 **
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_0=15m/s
v_f=0m/s
'ds=12 m
a=-10 m/s^2
vAve=(15m/s +0 m/s) / 2 =7.5 m/s
'ds=vAve*dt
=(7.5 m/s)*1.5 s
=11.25 m
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• 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:
12 m+11.25 m=23.25 meters
'ds=23.25 m
v_0=15 m/s
a= -10 m/s^2
v_f=v_0^2+2a*'dt
=sqrt(15 m/s)^2 + 2a * 'ds)
=sqrt (465 m^/s^2)
=21.6 m/s
dt=ds/vAve
vAve=(15 m/s+-21.6 m/s)/2
=-3.3 m/s
dt=ds/vAve
(-12 m)/(3.3 m/s)
=3.6 s
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• At what clock time(s) will the speed of the ball be 5 meters / second? answer/question/discussion: 'dt=(vf-v0)/a
• 'ds=(5 m/s-15 m/s)/-10 m/s=1 second
• 'dt=-5 m/s-15 m/s/-10 m/s/=2 second#$&*
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Good but be careful about grouping and units:
'ds=(5 m/s-15 m/s)/(-10 m/s^2)=1 second
'dt=-(5 m/s-15 m/s)/(-10 m/s^2)=2 seconds
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• At what clock time(s) will the ball be 20 meters above the ground?
• To achieve a 20 meter height the vertical position must change by 8 meters because we started at 12 m and we increase to 8 m.
v_f^2=v_0^2+2a*ds t
v_f= 8.1 m/s
'dt=dv/a=(-6.9 m/s)/-10 m/s^2=.69
Velocity= -8 m/s, 2.31 s
• How high will it be at the end of the sixth second?answer/question/discussion: ?????This problem confuses me, this is as far as I could get Help!
12 m = 15t + ½a * t^2
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This could work. a = -10 m/s^2.
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45 minutes
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For some reason, I am struggling on this part!!!! Ah!
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You're doing well, but check the discussion at
&#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
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