phy 201
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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:
15 m/s
• What will be its velocity at the end of two seconds?
answer/question/discussion:
5 m/s
• During the first two seconds, what therefore is its average velocity?
answer/question/discussion:
(5 + 25)/2 = 15
• How far does it therefore rise in the first two seconds?
answer/question/discussion:
‘ds = vAve(‘dt) = 15*2 = 30 cm
• What will be its velocity at the end of an additional second, and at the end of one more additional second?
answer/question/discussion:
-5 m/s
-15 m/s
• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion:
‘dt = ’dv / a = 25 / 10 = 2.5 s
‘ds = vAve(‘dt) = (25/2) * 2.5 = 31.25 cm
• 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:
‘dv = a * ‘dt = 10 * 4 = 40 cm/s
vf = v0 - ‘dv = 25 – 40 = -15 cm/s
vAve = (vf + v0)/2 = 25/2 = 12.5 cm/s
‘ds = ‘dt * vAve = 4 * 12.5 = 50 cm s2 = 50 – vAve = 50 – 31.25 = 18.75 cm
• How high will it be at the end of the sixth second?
answer/question/discussion: -10 cm
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20 min.
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Very good. No revision is required, but do check out the solution at the link below:
&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#