phy 121
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25 min.
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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 = +25m/s
vf = ?
a = -10 m/s^2
vf = 25m/s + -10 m/s^2 (1s)
vf = 25m/s + -10m/s
vf = 15m/s
Velocity will be 15m/s after 1 second.
• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
v0 = +25m/s
vf = ?
a = -10 m/s^2
vf = 25m/s + -10 m/s^2 (2s)
vf = 25m/s + -20m/s
vf = 5m/s
Velocity will be 5m/s after 2 seconds.
• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = (15m/s + 5m/s) / 2 = 10m/s
During the first two seconds velocity starts at 25 m/s and ends as 5 m/s
• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
20 meters (2s * 10m/s)
• What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> :
v0 = +25m/s
vf = ?
a = -10 m/s^2
vf = 25m/s + -10 m/s^2 (3s)
vf = 25m/s + -30m/s
vf = -5m/s
Velocity will be -5m/s after 3 seconds.
v0 = +25m/s
vf = ?
a = -10 m/s^2
vf = 25m/s + -10 m/s^2 (4s)
vf = 25m/s + -40m/s
vf = -15m/s
Velocity will be -15m/s after 4 seconds.
• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> :
When the ball reaches 0 (vf), it goes no further up. We need to 1st find the time it takes to reach that point.
v0 = +25m/s
vf = 0
a = -10 m/s^2
0 = 25m/s - 25m/s + -10 m/s^2 (`dt)
-25m/s / -10m/s^2 = -10m/s^2 / -10m/s^2 (`dt)
`dt = 2.5 seconds
To find the height:
`ds = (vf + v0) / 2 * `dt
`ds = (0 + 25m/s) / 2 * 2.5s
`ds = 12.5m/s * 2.5s
`ds = 31.25 meters
That height would be 31.25 meters at 2.5 seconds.
• 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: ->->->->->->->->->->->-> :
(15m/s + 5m/s – 5m/s – 15m/s) / 4 = 0
0 m/s at the end of 4 seconds, it has came back down to the starting point.
• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
v0 = +25m/s
vf = ?
a = -10 m/s^2
vf = 25m/s + -10 m/s^2 (6s)
vf = 25m/s + -60m/s
vf = -35m/s
Velocity will be -35m/s after 6 seconds.
To find the height:
`ds = (vf + v0) / 2 * `dt
`ds = (0 + 25m/s) / 2 * 6s
`ds = 12.5m/s * 6s
`ds = 75 meters
That original height up & down to point 0 is 62.5. So the height of the ball after 6 seconds would be (75m – 62.5m) = -12.5 meters.
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