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PHY 241
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: ->->->->->->->->->->->-> :
vf=?
v0=15m/s
'dt=?
`ds=?
a=-10m/s/s
vf will be 0 at highest point we are looking for dt and ds.
vf = v0 + a * `dt
0= 15m/s +-10m/s/s’dt
-15m/s=-10m/s/s’dt
`dt=1.5s
`ds = (15 + 0) / 2 * `1.5
`ds=11.25m but this doesn’t take into account that it started at 12 m so really it will rise to 23.25m above the ground
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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: ->->->->->->->->->->->-> :
it will strike the ground when `ds = -12 m/s
vf^2 = v0^2 + 2 a `ds.
vf^2 = 15^2 + 2 -10 (-12m/s)
vf^2 = 225+240
vf^2 = 465
vf=-21.6m/s
vf = v0 + a * `dt
-21.6m/s = 15m/s + -10m/s/s * `dt
-36.6m/s=-10m/s/s * `dt
Dt= 3.66s
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
5m/s = 15m/s + -10m/s/s * `dt
-10m/s=-10m/s/sdt
Dt=1s
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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: ->->->->->->->->->->->-> :
take into account we start at 12 m above the ground 20-12= 8m for ds
8m = 15m/s `dt + .5 -10m/s `dt^2
8m = 15m/s `dt -5m/s `dt^2
`dt= 0.69s and `dt= 2.3s
`ds = 15m/s 6s + .5 -10m/s/s 6^2s^2
`ds = 90m + -180
`ds=-90m this is below ground level so it will be at 0m
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