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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?
15m/s
I multiplied the downward acceleration (-10m/s^2) and the time (1s) and added this to the initial velocity of 25m/s which gave me a velocity of 15m/s.
What will be its velocity at the end of two seconds?
Using the same method in the first problem, I calculated the velocity after 2 seconds to be 5m/s.
During the first two seconds, what therefore is its average velocity?
15m/s
vAve = v0 +vf/2 = 25 + 5/2 = 15m/s
How far does it therefore rise in the first two seconds?
The ball rises 30m in the first two seconds.
`ds = vAve`dt = (15)(2) = 30m
What will be its velocity at the end of a additional second, and at the end of one more additional second?
At the end of an additional second, it will be -5m/s and at the end of one more additional second, -15m/s.
At what instant does the ball reach its maximum height, and how high has it risen by that instant?
The ball reaches its maximum height of 31.25m at 2.5s.
I used these formulas to make my calculations:
`dt = (vf - v0)/a and `ds = vf^2 - v0^2/2a
What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
vAve = (v0 + vf)/2 = (25m/s + -15m/s)/2 = 5m/s
`ds = vAve`dt = (5)(4) = 20m
How high will it be at the end of the sixth second?
vf = a`dt + v0 = (-10)(6) + 25 = -35m/s
vAve = (v0 + vf)/2 = (25) + (-35)/2 = -5m/s
`ds = vAve`dt = (-5m/s)(6s) = -30m or 0m (which means on the ground)
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35min.
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Please respond relative to goal of C.
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