phy121
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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?
Since the acceleration of gravity is downward, then an upward toss means the velocity is negative.
v0 + (a * `dt) = v
-25m/s + (10m/s/s * 1sec) = -15m/s
What will be its velocity at the end of two seconds?
Using the same equasion, we get:
-25m/s + (10m/s/s * 2sec) = -5m/s
During the first two seconds, what therefore is its average velocity?
(v0 + vF) / 2 = vAve
(-25m/s + -5m/s) / 2 = -15m/s
How far does it therefore rise in the first two seconds?
`dt * vAve = `ds
2sec * -15m/s = -30m
What will be its velocity at the end of a additional second, and at the end of one more additional second?
At one additional seconds (3 sec).
-25m/s + (10m/s/s * 3sec) = 5m/s
And at two additional seconds (4 sec).
-25m/s + (10m/s/s * 4sec) = 15m/s
At what instant does the ball reach its maximum height, and how high has it risen by that instant?
The point will be at v=0, so substituting it into the equasion, we get:
(v - v0) / a = `dt
(0 - (-25)) / 10 = 2.5sec
To find the `ds:
(v0 + vF) / 2 = vAve
(-25m/s + 0m/s) / 2 = -12.5m/s
vAve * `dt = `ds
-12.5m/s * 2.5sec = -31.25m
What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
velocity after four seconds:
-25m/s + (10m/s/s * 4sec) = 15m/s
Height after four seconds is:
vAve * `dt = `ds
[(-25m/s + 15m/s) / 2] * 4sec = -20m
How high will it be at the end of the sixth second?
-25m/s + (10m/s/s * 6sec) = 35m/s
vAve * `dt = `ds
[(-25m/s + 35m/s) / 2] * 6 = 30m
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20min
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&#Very good work. Let me know if you have questions. &#