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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?
o V= v0 +at
o V= 25 m/s + (-10 m/s^2) (1.00s)
o V= 25 m/s + (-10 m/s)
o V= 15 m/s
Acceleration was given a negative sign because it is being thrown upward. As the ball rises, its speed decrease until it reaches it highest point where its speed is zero. The ball then descends with an increasing speed.
What will be its velocity at the end of two seconds?
V= 25 m/s + (-10 m/s^2) ( 2.00 s)
V= 25 m/s + (-20 m/s)
V= 5 m/s
During the first two seconds, what therefore is its average velocity?
vAve = (15 m/s + 5 m/s) / 2 = 10 m/s
How far does it therefore rise in the first two seconds?
(10 m/s) ( 2.00s) = 20 meters
What will be its velocity at the end of an additional second, and at the end of one more additional second?
With an additional second:
V= 25 m/s + (-10 m/s) (3.00s)
V= -5m/s
With another second:
V= 25 m/s + (-10 m/s) ( 4.00s)
V= -15 m/s
At what instant does the ball reach its maximum height, and how high has it risen by that instant?
o By setting v= 0 at the max height and taking the equation v= v0 +at and solving for t we obtain the following:
o T= - v0/a = - 25 m/s / -10 m/s^s = 2.5 s
o Y= v^2 v^20 / 2 a = 0 (25 m/s)^2 / 2 (-10 m/s^2)
o Y= 31.25 m
What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
o vAve = 15 + 5 + 15 + 5 / 4 = 10 m/s
Acceleration is constant, so the v vs. t graph is linear. Therefore you need only average the initial and final velocities.
You get (25 m/s+ - 15 m/s) / 2 = 5 m/s.
o The result came out positive because the negative signs on the velocity tells if it is going upward or downward. Therefore, if I would have added the velocities together with the negative signs, the average velocity would come out to be 0 over the 4 seconds; this can not be true.
If you added up 25 + 15 + 5 + -5 + -15 you would get 25 m/s, which divided by the 5 velocities would give you 5 m/s.
o Y= v0t + 1/2at^2 = (25 m/s) (4.00s) + ½(-10 m/s)(4.00s)^2
o Y= 20 m
How high will it be at the end of the sixth second?
o If the ball is at its max height at 2.5 s, which is the half the time it takes the ball to go up and fall back to it original position, it would take the same time to reach the maximum height as to fall back to the starting point. Therefore, the ball would be back in the persons hand at 5 s. Thus the ball will be at rest at 6s.
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1 hour
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I'm not quite sure about the average velocity after 4 seconds. I hope I did it right!
&#Your work looks good. See my notes. Let me know if you have any questions. &#