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Phy 121
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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: ->->->->->->->->->->->-> :
v0 = 25 m/s a = -10 m/s^2 t = 1s
a = (vf - v0)/'dt vf = a * 'dt + v0
vf = -10 m/s^2 * 1s + 25 m/s = 15 m/s
vf in this case is not final velocity, but it is the velocity after 1 second
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = -10 m/s^2 * 2s + 25 m/s = 5 m/s
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
(vf + v0) / 2 = (25 m/s + 5 m/s) / 2 = 15 m/s = vAve
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
'ds = vAve * 'dt 'ds = 15 m/s * 2 sec
'ds = 30 meters
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What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> :
I will use the same reasoning I have been for these questions.
After 3 seconds: vf = -10 m/s^2 * 3s + 25 m/s = -5 m/s
After 4 seconds: vf = -10 m/s^2 * 4s + 25 m/s = -15 m/s
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At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> :
It reaches a maximum height when it's velocity changes from positive to negative (or at 0 m/s) so:
0 m/s = -10 m/s^2 * t + 25 m/s t = -25 m/s / -10 m/s^s = 2.5 s
'ds = vAve * 'dt We must first find vAve at 2.5 seconds. vf = -10 m/s^2 * 2.5 + 25 m/s In this case it is 'ds = (25 m/s + 0 m/s) / 2 * 2.5 = 31.25 meters.
I could also use 'ds = v0 * 'dt + .5 * a * 'dt^2, 'ds = 25 m/s * 2.5 s + .5 * -10 m/s^2 * 2.5^2 = 31.25 meters.
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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: ->->->->->->->->->->->-> :
v0 = 25 m/s, v1 = 15m/s, v2 = 5 m/s, v3 = -5 m/s, v4 = -15 m/s
vAve = (25 m/s + 15 m/s + 5 m/s - 5 m/s - 15 m/s) / 5 = 5 m/s
'ds = vAve * 'dt, 'ds = 5 m/s * 4 sec = 20 meters.
**This does not seem right to me**
The ball travels 25 meters, then 15 meters, then 5 meters, then it drops 5 meters, and then drops 15 meters, which adds up to 25 meters.
I could also use 'ds = v0 * 'dt + .5 * a * 'dt^2, 'ds = 25 m/s * 4 s + .5 * -10 m/s^2 * 4^2 = 20 meters.
**I guess 20 meters is correct**
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
'ds = v0 * 'dt + .5 * a * 'dt^2
'ds = 25 m/s * 6 s + .5 * -10 m/s^2 * 6^2 = -30 meters.
I would think that it would have reached its initial takeoff point, so if it bounces it would have positive 'ds, if it landed and stuck it would have 'ds = 0 meters, and if it free falls then it would have 'ds = -30 meters.
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15 min
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