Assignment 8

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18:15:48 `q001. The accepted value of the acceleration of gravity is approximately 980 cm/s^2 or 9.8 m/s^2. This will be the acceleration, accurate at most places within 1 cm/s^2, of any object which falls freely, that is without the interference of any other force, near the surface of the Earth. If you were to step off of a table and were to fall 1 meter without hitting anything, you would very nearly approximate a freely falling object. How fast would you be traveling when you reached the ground?

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RESPONSE --> I don;t undersatnd the example in the book....where do I get the time it would take to fall?

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18:20:27 You would have an initial vertical velocity of 0, and would accelerate at 9.8 m/s^2 in the same direction as your 1 meter vertical displacement. You would also have a slight horizontal velocity (you don't step off of a table without moving a bit in the horizontal direction, and you would very likely maintain a small horizontal velocity as you fell), but this would have no effect on your vertical motion. So your vertical velocity is a uniform acceleration with v0 = 0, `ds = 1 meter and a = 9.8 m/s^2. The equation vf^2 = v0^2 + 2 a `ds contains the three known variables and can therefore be used to find the desired final velocity. We obtain vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m)= +- `sqrt ( 19.6 m^2 / s^2) = +- 4.4 m/s, approx. Since the acceleration and displacement were in the direction chosen as positive, we conclude that the final velocity will be in the same direction and we choose the solution vf = +4.4 m/s.

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RESPONSE --> Ok...I can work it out looking at the example...myproblem is still really undersatnd what everything stands for and how to work the equations withour referring to something

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18:33:44 `q002. If you jump vertically upward, leaving the ground with a vertical velocity of 3 m/s, how high will you be at the highest point of your jump? Note that as soon as you leave the ground, you are under the influence of only the gravitational force. All the forces that you exerted with your legs and other parts of your body to attain the 3 m/s velocity have done their work and are no longer acting on you. All you have to show for it is that 3 m/s velocity. So as soon as you leave the ground, you begin experiencing an acceleration of 9.8 m/s^2 in the downward direction. Now again, how high will you be at the highest point of your jump?

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RESPONSE --> okay I think I did this one...v =0, a = -9.80m/s dt= 3m/m 0-(3m/s)2 / 2 (-9.80m/s^2 = 9m/s / 19.6 m/s = .46m

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18:37:01 From the instant the leave the ground until the instant you reach your highest point, you have an acceleration of 9.8 m/s^2 in the downward direction. Since you are jumping upward, and since we can take our choice of whether upward or downward is the positive direction, we choose the upward direction as positive. You might have chosen the downward direction, and we will see in a moment how you should have proceeded after doing so. For now, using the upward direction as positive, we see that you have an initial velocity of v0 = + 3 m/s and an acceleration of a = -9.8 m/s^2. In order to use any of the equations of motion, each of which involves four variables, you should have the values of three variables. So far you only have two, v0 and a. {}What other variable might you know? If you think about it, you will notice that when objects tossed in the air reach their highest point they stop for an instant before falling back down. That is precisely what will happen to you. At the highest point your velocity will be 0. Since the highest point is the last point we are considering, we see that for your motion from the ground to the highest point, vf = 0. Therefore we are modeling a uniform acceleration situation with v0 = +3 m/s, a = -9.8 m/s^2 and vf = 0. We wish to find the displacement `ds. Unfortunately none of the equations of uniformly accelerated motion contain the four variables v0, a, vf and `ds. This situation can be easily reasoned out from an understanding of the basic quantities. We can find the change in velocity to be -3 meters/second; since the acceleration is equal to the change in velocity divided by the time interval we quickly determine that the time interval is equal to the change in velocity divided by the acceleration, which is `dt = -3 m/s / (-9.8 m/s^2) = .3 sec, approx.; then we multiply the .3 second time interval by the 1.5 m/s average velocity to obtain `ds = .45 meters. However if we wish to use the equations, we can begin with the equation vf = v0 + a `dt and solve to find `dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec. We can then use the equation `ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m. This solution closely parallels and is completely equivalent to the direct reasoning process, and shows that and initial velocity of 3 meters/second should carry a jumper to a vertical height of .45 meters, approximately 18 inches. This is a fairly average vertical jump. If the negative direction had been chosen as positive then we would have a = +9.8 m/s^2, v0 = -3 m/s^2 (v0 is be in the direction opposite the acceleration so if acceleration is positive then initial velocity is negative) and again vf = 0 m/s (0 m/s is the same whether going up or down). The steps of the solution will be the same and the same result will be obtained, except that `ds will be -.45 m--a negative displacement, but where the positive direction is down. That is we move .45 m in the direction opposite to positive, meaning we move .45 meters upward.

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RESPONSE --> That took a lot of thinking and I had to refer to the book...but i still didn't solve the way this says to solve...where did the 1.5 m/s Avel come from?

Initial velocity is 0, final velocity is 3 m/s and acceleration is uniform, so vAve is the average of initial and final velocities.

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18:54:24 `q003. If you roll a ball along a horizontal table so that it rolls off the edge of the table at a velocity of 3 m/s, the ball will continue traveling in the horizontal direction without changing its velocity appreciably, and at the same time will fall to the floor in the same time as it would had it been simply dropped from the edge of the table. If the vertical distance from the edge of the table to the floor is .9 meters, then how far will the ball travel in the horizontal direction as it falls?

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RESPONSE --> I cam it with .54 m but I don't know that I used the right equation

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18:59:37 A ball dropped from rest at a height of .9 meters will fall to the ground with a uniform vertical acceleration of 9.8 m/s^2 downward. Selecting the downward direction as positive we have `ds = .9 meters, a = 9.8 m/s^2 and v0 = 0. Using the equation `ds = v0 `dt + .5 a `dt^2 we see that v0 = 0 simplifies the equation to `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec, approx.. Since the ball rolls off the edge of the table with only a horizontal velocity, its initial vertical velocity is still zero and it still falls to the floor in .42 seconds. Since its horizontal velocity remains at 3 m/s, it travels through a displacement of 3 m/s * .42 sec = 1.26 meters in this time.

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RESPONSE --> Okay...I see how it is done but I just don't undersatnd why we do all these steps ...can't I just out with sqrt 2.9m/9.8m/s^2 and keep going from there...

You have four equations of motion, and everything you do when you use equations should start from and come back to that set of four equations.

The equation that applies here is `ds = v0 `dt + .5 a `dt^2.

Make sure you know those four equations thoroughly, and how to solve them for the various variables.

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See my notes and answers to your questions. Let me know if you have additional questions.