Your 'cq_1_8.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
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?
&&&&t = (v – v0)/a = (0-25m/s^2)/-10m/s^2 = 2.5s, it reaches its max height after 2.5 seconds
y = (v^2 – v0^2)/ (2*a) = ( 0 – (25m/s)^2)/(2*-10m/s^2) = -625m^2/s^2 / -20m/s^2 = 31.25m
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?
&&&&vAve = (15m/s + 5m/s + -5m/s + -15m/s)/4 = 0
You can get in trouble using the intermediate velocities, which in a trapezoidal approximation would each get double the weight of the endpoint velocities.
You also haven't included the initial velocity in your average.
Since the acceleration is uniform, the average velocity is the average of initial and final velocities: vAve = (25 m/s + (-15 m/s) ) / 2 = 5 m/s.
&&&&The ball reached the max height after 2.5secs, therefore after 2.5 secs it will be in a downward descent. I calculated the height after 4 secs by calculating the height 1.5 secs after the max height. V= v0 + at = 0 + -10m/s^2 * 1.5s = 15m/s, this is the velocity at 4 secs. Now using v^2 = v0^2 + 2ay, rewriting it we have y = (v^2 – v0^2)/2a, substituting v=15m/s^2, a=10m/s^2, and v0 = 0, then y = (225m^2/s^2)/20m/s^2 = 11.25m, therefore the ball is 11.25m below the maximum height. So the ball is at height = 31.25m – 11.25m = 20m
Good, but be sure you understand that you don't need to calculate any of the intermediate velocities. The ball avearges 5 m/s for 4 s so its displacement during that 4 s is +20 m.
?How high will it be at the end of the sixth second?
&&&&By common sense the ball should be on the ground, it reaches at max height at 2.5 secs, it will take 2.5 secs to make it back to the starting position. That’s 5 secs, 6 seconds would mean it has already landed. I need more information to solve this problem. The question doesn’t state the height above the ground at which the ball was thrown.
The correct solution would assume uniform acceleration for the entire 6 seconds, then note that if something interferes with the uniformity of the acceleration, the solution won't necessarily correspond to reality.
The velocity after 6 sec is -35 m/s so ave vel for the six sec is -5 m/s and displacement is -30 m. If the ball is thrown from a point 30 m high or more, this solution might be valid. There might also be a hole in the ground.
At those speeds air resistance is likely to be significant so the solution won't be completely accurate.
** **
50 mins
** **
This is a resubmission, could you check the last 3 questions to make sure I am on the right track.
&#This looks good. See my notes. Let me know if you have any questions. &#