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course Phy 121
20/Jul/2010 20:52
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:
v = v0 + a * 'dt
v = 25 m/s - 10 m/s^2 * 1 s
v = 15 m/s
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What will be its velocity at the end of two seconds?
answer/question/discussion:
v = v0 + a * 'dt
v = 25 m/s - 10 m/s^2 * 2 s
v = 5 m/s
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion:
v_Ave = (vf + v0) /2
v_Ave = 25 m/s / 2 = 12.5 m/s
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How far does it therefore rise in the first two seconds?
answer/question/discussion:
'ds = s0 + .5 * (vf + v0) * 'dt
'ds = 0 + .5 * 25 m/s * 2 s
'ds = 25 m
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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:
v = v0 + a * 'dt
v = 25 m/s - (10 m/s^2 * 3 s)
v = -5 m/s
v = v0 + a * 'dt
v = 25 m/s - (10 m/s^2 * 4 s)
v = -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:
v = v0 + a * 'dt
v = 25 m/s - (10 m/s^2 * 2.5 s)
v = 0 m/s
'ds = s0 + .5 * (vf + v0) * 'dt
'ds = 0 + .5 * 25 m/s * 2.5 s
'ds = 31.25 m
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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:
v = v0 + a * 'dt
v = 25 m/s - 10 m/s^2 * 4 s
v = -15 m/s
v_Ave = vf + v0 /2
v_Ave = 0 m/s + 25 m/s / 2 = 12.5 m/s
'ds = s0 + .5 * (vf + v0) * 'dt
'ds = 0 + .5 * 25 m/s * 2.5 s
'ds = 31.25 m
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How high will it be at the end of the sixth second?
'ds = s0 + .5 * (vf + v0) * 'dt
'ds = 0 + .5 * 25 m/s * 6 s
'ds = 31.25 m
answer/question/discussion:
Total time in air:
y = v0 * t + .5 * -9.8 m/s^2 * t^2
y = 25 m/s^2 / 4.9 m/s^2 = 5.0 s
Ball would return to origin @5.0 s
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&#Very good responses. Let me know if you have questions. &#
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