course Phy201
I don't think that I am doing this right, so I'm not going to do anymore of Assignment 8 until I receive something back from you.
Seed Question 8.1A 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?
With v0 = 25m/s, `dt = 1s, and a = -10m/s^2. Then, a = `dv/`dt, -10m/s^2 = (vf -25m/s)/1s
Multiplying both sides by 1s yields, -10m/s = vf ?25m/s. Adding 25m/s to both sides we see vf =15m/s. The velocity after 1 second is 15m/s.
Good use of the equations and definitions.
An acceleration of - 10 m/s^2 means that the velocity changes by -10 m/s every second. So this result could have been reasoned out directly and intuitively. It's imporant that you understand it both ways.
?What will be its velocity at the end of two seconds?
Using the same equation above, substituting 2s for `dt, we get vf = 5m/s
again, no need for an equation, but the equation did give you the correct result
?During the first two seconds, what therefore is its average velocity?
vAve = (v0 + vf)/2 = 30m/s / 2 = 15m/s
?How far does it therefore rise in the first two seconds?
15m/s * 2s = 30m
?What will be its velocity at the end of a additional second, and at the end of one more additional second?
Vf = 25m/s ?10m/s^2*3s = -5m/s, vf = 25m/s -10m/s^2 * 4s = -15m/s
everything is correct up to this point
?At what instant does the ball reach its maximum height, and how high has it risen by that instant?
31.9m
allowing for a little roundoff this is correct. How did you get this result?
?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:
?How high will it be at the end of the sixth second?
answer/question/discussion:
Nearly everything is done correctly. Very good use of equations, but be sure you understand this intuitively as well.
See my notes and submit a copy of this document with your solutions to the final two problems inserted and indicated by &&&&. As always questions are welcome.