course Phy 201 ÁKöø¶¹üV©UÙ¹«ÝÝî¹zØøÛúŽPÔç•Õassignment #021
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00:54:18 `q001. Note that this assignment contains 3 questions. . A projectile has an initial velocity of 12 meters/second, entirely in the horizontal direction. After falling to a level floor three meters lower than the initial position, what will be the magnitude and direction of the projectile's velocity vector?
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RESPONSE --> x component=12m/s ycomponent= acceleration of -9.8m/s^2 `ds=-3m vf^2=v0^2 +2a(`ds) vf^2=0 +2(9.8m/s^2)(-3m) vf^2=58.8m^2/s^2 vf=+-7.67m/s x^2 +y^2=z^2 144m^2/s^2 +58.8m^2/s^2=z^2 z^2=202.8m^2s^2 z=14.24m/s arctan of y/x= arctan(7.67/12) arctan(.6392)=32.59 degrees confidence assessment: 1
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00:56:45 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 --> Because the velocity is directed downward=-7.7m/s Should have been arctan y/x=-7.7m/s/12m/s=-35 degrees yis neg so add 360 degrees= 325 degrees self critique assessment: 2
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01:15:58 `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 --> v0x=20m/s*cos30deg=17.32m/s v0y=20m/s*sin30deg=9.98m/s (v0x)^2 + (v0y)^2=(v0z)^2 (17.32m/s)^2 + (9.98m/s)^2=z^2 400m^2/s^2=z^2 20=z for vertical= vf^2=v0^2 +2a(`ds) vf^2=0 + 2(9.8m/s^2)(12m)
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01:23:58 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 --> Am confused as to why you only need the vertical motion.
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01:37:10 `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 --> Should you use the `dt and velocities from earlier? aave=`dv/`dt `dv=28.3m/s `dt=2.7s 28.3m/s/2.7s=10.48m/s^2 vf^2=v0^2+2a(`ds) (-18.3m/s)^2=(10m/s)^2 + 2(10.48m/s^2)(`ds) 334.89m^2/s^2=100m^2/s^2 + 20.96m/s^2(`ds) Subtract 100m^2/s^2 234.89m^2/s^2=20.96m/s^2(`ds) Divide by 20.96m/s^2 11.21m=`ds confidence assessment: 1
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01:38:44 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 --> Horizontal velocity=20m/s *cosine(30degrees)=17.3m/s Should calculate the horizontal velocity and use the `dt from earlier 17.3m/s for 2.7s= 46m self critique assessment: 2
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