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Assn 8 Seed 1
A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.
What will be the velocity of the ball after one second?
answer/question/discussion:
vf= v0 + a * `dt, so vf = 25 + (-10)1= 25-10= 15m/s
What will be its velocity at the end of two seconds?
answer/question/discussion:
vf= v0 + a * `dt, so vf = 25 + (-10)2= 25-20= 5m/s
During the first two seconds, what therefore is its average velocity?
answer/question/discussion:
(v0+vf)/2= vAve, so vAve = (25+5)/2= 30/2= 15m/s
How far does it therefore rise in the first two seconds?
answer/question/discussion:
`ds/`dt= vAve, or `ds = vAve * `dt, so `ds = 15m/s*2s= 30m
What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion:
vf= v0+ a *`dt, so after 3 seconds, vf= 25 + (-10)*3= 25-30= -5m/s
vf= v0+ a *`dt, so after 4 seconds, vf= 25 + (-10)*4= 25-40= -15m/s
At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion:
(vf^2-v0^2)/2a= `ds, (0^2-25^2)/2(-10)= -625/-20= 31.25m
What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
answer/question/discussion:
(vf+v0)/2= vAve, so (-15+25)/2= 10/2= 5m/s
How high will it be at the end of the sixth second?
answer/question/discussion:
`ds= v0`dt+.5a*`dt^2, so `ds= 25*6 + .5(-10)6^2= 150 + (-5)36= 150-180= -30m
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17min
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