#$&*
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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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: ->->->->->->->->->->->-> :
v0=15
'ds= 12
a=10m/s^2
vf=0
0=15-10'dt
10'dt=15
'dt=15/10
'dt=1.5 seconds to get to its highest point
1.5*15=22.5 meters into the air or 12+22.5=35.5 meters from the position it was thrown from.
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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: ->->->->->->->->->->->-> :
'ds=22.5
vf=v0+a'dt
vf=15-10*3
vf=-15m/s
'dv=0
this is velocity when it reaches its original 12 m position, not when it reaches the ground
I got lost in here somewhere
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
'dt=5-1.5= 3.5
vf=15+10(3.5)
vf=50m/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: ->->->->->->->->->->->-> :
just over a second
vf=15+10(4.5)
vf=60m/s
I think it will hit the ground
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15 minutes
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You understand but you're missing a few of the details. Be sure you understand the discussion at the link. If not, send a revision with attempted solutions and/or questions. Otherwise, no need.
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Then please compare your solutions with the expanded discussion at the link
Solution
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