course Phy 201 xzf蕾lĔ嵳assignment #021
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10:08:23 Explain how to obtain the final speed and direction of motion of a projectile which starts with known velocity in the horizontal direction and falls a known vertical distance, using the analysis of vertical and horizontal motion and vectors.
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RESPONSE --> You can take the final and initial velocities and then find there difference and multiply by the motion to get the speed and then you can use cos and sin to find the direction.
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10:09:09 ** The horizontal velocity is unchanging so the horizontal component is always equal to the known initial horizontal velocity. The vertical velocity starts at 0, with acceleration thru a known distance at 9.8 m/s^2 downward. The final vertical velocity is easily found using the fourth equation of motion. We therefore know the x (horizontal) and y (vertical) components of the velocity. Using the Pythagorean Theorem and arctan (vy / vx) we find the speed and direction of the motion. **
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RESPONSE --> I understand that the vertical velocity starts at 0, with acceleration thru a known distance at 9.8 m/s^2 downward. The final vertical velocity is easily found using the fourth equation of motion and therefore know the x (horizontal) and y (vertical) components of the velocity. Using the Pythagorean Theorem and arctan (vy / vx) we find the speed and direction of the motion
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10:13:53 Give at least three examples of vector quantities for which we might wish to find the components from magnitude and direction. Explain the meaning of the magnitude and the direction of each, and explain the meaning of the vector components.
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RESPONSE --> 5,6 ; 7,8 ; 9,10 You would use the pythagorem therom to find the c2 component, then once you have the c component you have your magnitude and then you can take the sin and cos of the c^2 to find the direction of each vector quantity.
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10:14:19 ** GOOD STUDENT RESPONSE: Examples might include: A force acting on an object causing it to move in an angular direction. A ball falling to the ground with a certain velocity and angle. A two car collision; velocity and momentum are both vector quantities and both important for analyzing the collision.. The magnitude and directiohn of the relsultant is the velocity and direction of travel. The vector components are the horizontal and vertical components that would produce the same effect as the resultant.
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RESPONSE --> I understand that if a force acting on an object causing it to move in an angular direction. A ball falling to the ground with a certain velocity and angle. A two car collision; velocity and momentum are both vector quantities and both important for analyzing the collision.. The magnitude and directiohn of the relsultant is the velocity and direction of travel. The vector components are the horizontal and vertical components that would produce the same effect as the resultant.
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course Phy 201 Sorry, the last assignment 21 that I sent was incorrect, this is the correct assignment 21. MǷnj|lassignment #021
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RESPONSE --> Using phy. therom, I used 12m/s^2 + 9m/s^2 = 225 m/s^2 and the square root of this answer is 15m/s so therefore, the magnitude is 15m/s and the velocity vector will be 180degrees. confidence assessment: 1
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22:47:35 To answer this question we must first determine the horizontal and vertical velocities of the projectile at the instant it first encounters the floor. The horizontal velocity will remain at 12 meters/second. The vertical velocity will be the velocity attained by a falling object which is released from rest and allowed to fall three meters under the influence of gravity. Thus the vertical motion will be characterized by initial velocity v0 = 0, displacement `ds = 3 meters and acceleration a = 9.8 meters/second ^ 2. The fourth equation of motion, vf^2 = v0^2 + 2 a `ds, yields final vel in y direction: vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7 meters/second. Since we took the acceleration to be in the positive direction the final velocity will be + 7.7 meters/second. This final velocity is in the downward direction. On a standard x-y coordinate system, this velocity will be directed along the negative y axis and the final velocity will have y coordinate -7.7 m/s and x coordinate 12 meters/second. The magnitude of the final velocity is therefore `sqrt((12 meters/second) ^ 2 + (-7.7 meters/second) ^ 2 ) = 14.2 meters/second, approximately. The direction of the final velocity will therefore be arctan ( (-7.7 meters/second) / (12 meters/second) ) = -35 degrees, very approximately, as measured in the counterclockwise direction from the positive x axis. The direction of the projectile at this instant is therefore 35 degrees below horizontal. This angle is more commonly expressed as 360 degrees + (-35 degrees) = 325 degrees.
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RESPONSE --> I understand that v0 = 0, displacement `ds = 3 meters and acceleration a = 9.8 meters/second ^ 2 and that using the 4th equation of motion, vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7 meters/second and since a was positive, then the vf will be positive as well. Therefore, on the x y coordniate system, the velocity will be along he negative y axis and the final velocity will have y coordinate -7.7 m/s and x coordinate 12 meters/second, and to find the magnitude you simply take sqrt((12 meters/second) ^ 2 + (-7.7 meters/second) ^ 2 ) = 14.2 meters/second and the direction of the final velocity will therefore be arctan ( (-7.7 meters/second) / (12 meters/second) ) = -35 degrees and then you add 360 degrees to that which will equal 325degrees. self critique assessment: 2
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23:04:43 `q002. A projectile is given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal, at an altitude of 12 meters above a level surface. How long does it take for the projectile to reach the level surface?
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RESPONSE --> I used the equation, vf^2= v0^2 + 2a ds therefore, I plugged in, vf^2 = 20m/s^2 + 2 9.8m/s^2 * 12m = vf = 25.2m/s then I took dt = ds/vAve and therefore, dt = 12m/ (20m/s + 25.2 m/s / 2 ) dt = .53sec confidence assessment: 1
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23:10:00 To determine the time required to reach the level surface we need only analyze the vertical motion of the projectile. The acceleration in the vertical direction will be 9.8 meters/second ^ 2 in the downward direction, and the displacement will be 12 meters in the downward direction. Taking the initial velocity to be upward into the right, we situate our x-y coordinate system with the y direction vertically upward and the x direction toward the right. Thus the initial velocity in the vertical direction will be equal to the y component of the initial velocity, which is v0y = 20 meters/second * sine (30 degrees) = 10 meters/second. Characterizing the vertical motion by v0 = 10 meters/second, `ds = -12 meters (`ds is downward while the initial velocity is upward, so a positive initial velocity implies a negative displacement), and a = -9.8 meters/second ^ 2, we see that we can find the time `dt required to reach the level surface using either the third equation of motion `ds = v0 `dt + .5 a `dt^2, or we can use the fourth equation vf^2 = v0^2 + 2 a `ds to find vf after which we can easily find `dt. To avoid having to solve a quadratic in `dt we choose to start with the fourth equation. We obtain vf = +-`sqrt ( (10 meters/second) ^ 2 + 2 * (-9.8 meters/second ^ 2) * (-12 meters) ) = +-18.3 meters/second, approximately. Since we know that the final velocity will be in the downward direction, we choose vf = -18.3 meters/second. We can now find the average velocity in the y direction. Averaging the initial 10 meters/second with the final -18.3 meters/second, we see that the average vertical velocity is -4.2 meters/second. Thus the time required for the -12 meters displacement is `dt = `ds / vAve = -12 meters/(-4.2 meters/second) = 2.7 seconds.
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RESPONSE --> I understand that v0y = 20 meters/second * sine (30 degrees) = 10 meters/second and that ds = -12m and a = -9.8m/s^2 , and that vf= +-`sqrt ( (10 meters/second) ^ 2 + 2 * (-9.8 meters/second ^ 2) * (-12 meters) ) = +-18.3 meters/second and therefore the average vertical velocity = -4.3m/s and dt = ds/vAve and therefore, dt= -12m/ -4.2m/s and dt = 2.7sec self critique assessment: 2
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23:20:03 `q003. What will be the horizontal distance traveled by the projectile in the preceding exercise, from the initial instant to the instant the projectile strikes the flat surface.
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RESPONSE --> I used the equation squareroot 2ds/a therefore, I took the squareroot of 2 * 12m/s / 9.8m/s^2 therefore, it equals 1.56m/s then I took this times the velocity which was 20m/s = 31.2m/s confidence assessment: 1
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23:21:52 The horizontal velocity of the projectile will not change so if we can find this horizontal velocity, knowing that the projectile travels for 2.7 seconds we can easily find the horizontal range. The horizontal velocity of the projectile is simply the x component of the velocity: horizontal velocity = 20 meters/second * cosine (30 degrees) = 17.3 meters/second. Moving at this rate for 2.7 seconds the projectile travels distance 17.3 meters/second * 2.7 seconds = 46 meters, approximately.
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RESPONSE --> I understand that the horizontal velocity = 20 meters/second * cosine (30 degrees) = 17.3 meters/second and that moving at this rate for 2.7 seconds the projectile travels distance 17.3 meters/second * 2.7 seconds = 46 meters self critique assessment: 2
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