phy201
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.
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8.2
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: ->->->->->->->->->->->-> :
'ds= 12meters
V0= 15m/s
a =10m/s^2
Vf^2 = V0^2 + 2a * 'ds
Vf^2= (15m/s)^2 + 2(10m/s^2) * 12meters
Vf^2 = 225m^2/s^2 + 20m/s^2 * 12 meters
Vf^2= 225m^s/s^2 + 240m^2/s^2
Vf^2= 465m^2/s^2
sqrt vf^2= sqrt 465m^2/s^2
vf= 22m/s
Vf= v0 + a * 'dt
22m/s = 15m/s + 10m/s^2 * 'dt
7m/s = 10m/s^2 *'dt
'dt = .7 seconds
'ds = vave * 'dt
Vave= (15m/s + 22m/s) / 2
vave= 18.5m/s
'ds = 18.5m/s * .7sec
'ds = 13 meters
After .7 of a second the ball has gone 13 meters
#$&*
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike
the ground?
answer/question/discussion: ->->->->->->->->->->->-> :
'ds = 13 meters (total to hit ground)
V0 = 15m/s
a = 10 m/s^2
I think we need vf
vf^2 = V0^2 + 2a * 'ds
Vf^2 = 15m/s + 2(10m/s) * 13meters
vf^2= 15m^2/s^2 + 20m/s * 13meters
Vf^2 = 15m^2/s^2 + 260m^2/s^2
Sqrt Vf^2= sqrt 275m^2/s^2
vf= 17m/s
vave= 16m/s
'ds = vave * 'dt
13meters = 16m/s * 'dt
'dt = .8 seconds
So if the ball start at an initial velocity of 15m/s and has gone up a total of 13 meters when it travels
back down the 13 meters at an acceleration consistent with gravity pulling at 10m/s^2 will be at a speed of
17m/s for .8 of a second.
#$&*
At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
Vf= 5 m/s
a= 10m/s^2
ds= 13 meters
Vf^2 = V0^2 + 2a * 'ds
(5m/s)^2= V0^2 + 2(10m/s^2) * 13meters
25m^/s^2 = V0^2 + 20 m/s^2 *13m
25m^2/s^2 = V0^2 + 260m^2/s^2
V0^2 = 235m^2/s^2
V0 = 15m/s
'ds = vave * 'dt
13m = 10m/s * 'dt
'dt = 1.3 seconds
#$&*
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: ->->->->->->->->->->->-> :
'ds = 20 meters
V0 = 15m/s
a = 10m/s^2
Vf^2 = V0^2 + 2a * 'ds
Vf^2 = (15m/s)^2 + 2(10m/s^2) * 20meters
Vf^2 = 225m^2/s^2 + 400m^2/s^2
Vf= 25m/s
vave= (25m/s + 15m/s) / 2
vave= 20m/s
ds= vave * 'dt
20meters = 20m/s * dt
dt= 1 second
'ds= vave * 6seconds
'ds = 20m/s * 6 seconds
'ds = 120 meters
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about an 30 min to an hour
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You're doing really well identifying quantities and using the equations. All you really need to do to get these problems is choose your positive direction, and give the various quantities + or - signs accordingly.
&#Please compare your 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. &#