course Phy 231 zÖøÈ´ðýݘ˜ïÞìÊüæ—¼óžºä÷^zc|”assignment #008
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15:46:35 `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 --> From the problem we know that `ds = 1m and since it's a free fall problem, the acceleration is the gravity constant 9.8m/s^2. My initial velocity at the top of the table is 0 m/s. The motion equations have four variables, we know 3 of them and we know that we're looking for the final velocity so we can use the equation: Vf^2 = Vo^2 + 2a`ds Vf^2 = 0 + 2(9.8m/s)(1m) Vf^2 = 19.6m^2/s Vf = +- 4.4m/s Since I would be falling in the downard direction, we will take that to be the positive direction. The acceleration is moving in the same way, so the velocity would as well. This means we would take the positive value from the square root so the final velocity = +4.4m/s confidence assessment: 3
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15:46:39 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 self critique assessment: 2
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15:49:59 `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 --> If the inital velocity starts when I leave the ground, then Vo = +3m/s. Since I'm moving in the upward direction, I will choose the positive position to be in the upward direction. If gravity is acting down on me, this gives me a negative acceleration of -9.8m/s^2. Whatever goes up must come down, and at my maximum height the velocity changes from + to -, meaning the point where it changes the final velocity is 0. Since we don't know a time interval we will use the equation Vf^2 = Vo^2 + 2a`ds and solve for `ds. 0 = 3^2 + 2(-9.8m/s^2)`ds 0 = 9 - 19.6m/s^2 `ds 19.6`ds = 9 `ds = 0.46m Therefore my maximum height will be attained at 0.46m confidence assessment: 3
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15:50:03 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 --> OK self critique assessment: 2
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15:58:29 `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 --> The initial vertical velocity of the ball will be 0. After it hits the ground it will continue to roll so its final horizontal velocity will remain at 3 m/s. We will take the downward direction to be positve. It's vertical displacement is 0.9m and it's vertical acceleration is 9.8m/s^2. We need to solve for its vertical final velocity so we use the equation: Vf^2 = Vo^2 + 2a`ds Vf^2 = 0 + 2(9.8m/s)(0.9m) Vf^2 = 17.64m^2/s Vf = 4.2 m/s. By finding the final velocity in the vertical direction, we can now find `dt. Vf = Vo + a`dt 4.2 = 0 + 9.8`dt `dt = 0.43s Since it keeps it's horizontal velocity of 3 m/s, we multiply this by the time interval to get how far it traveled in the horizontal directino. 3 m/s * 0.43s = 1.29m The ball travels 1.29m in the horizontal direction during its fall. confidence assessment: 3
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15:58:50 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 --> I rounded differently, but the got the answer the same way. self critique assessment: 2
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