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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: ->->->->->->->->->->->-> sion:
v0=15m/s
vf=0 m/s
vAve= 15m/s /2= 7.5 m/s
'dv= 0m/s-15m/s= -15m/s
a= -10 m/s^2
'dt= 'dv/a= -15m/s/-10m/s= 1.5s
'ds= vAve*'dt= 7.5 m/s * 1.5s= 11.25 m
11.25 meters is the highest point after 1.5s.
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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: ->->->->->->->->->->->-> sion:
'ds=-12m
vf^2 = v0^2 + 2 a `ds
vf= sqrt (v0^2 + 2 a `ds)
vf= sqrt((15m/s)^2 + 2 * 10 m/s^2 * 12m
vf= sqrt(225m^2/s^2)+ 240m^2/s^2)
vf=-21.56 m/s
Since this is for the total 'ds, vAve=(15m/s+-21.56m/s) /2
vAve=vf at grouund=3.28 m/s
'dt= `ds/vAve
'dt= -12m/3.28 m/s = 3.66 s
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf-v0 / a = (5m/s - 15m/s) / -10m/s^2 = 1 s
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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: ->->->->->->->->->->->-> sion:
Stating at 12 meters 20 meters will be 8 meters from inital point. 'ds=8
vf^2 = v0^2 + 2 a `ds
vf= sqrt (v0^2 + 2 a `ds)
vf= sqrt((15m/s)^2 + 2 * -10 m/s^2 * 8m
vf= sqrt(225m^2/s^2)+ -160m^2/s^2)
vf=-8.06 m/s
'dt= 'dv/a= -6.94m/s/-10m/s= .694s
The ball would have hit the ground after 3.6s, so after 6 seconds
ds= vAve*'dt= 3.28 m/s * 6s=19.28m
19.28m-12m=7.28m below ground.
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&#Very good responses. Let me know if you have questions. &#