course Phy 231
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17:20:37 `q001. Note that this assignment contains 4 questions. . How long does it take for an object dropped from rest to fall 2 meters under the influence of gravity?
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RESPONSE --> v0 = 0 m/s d = 2m a = 9.8 m/s^2 d = v0t + 1/2 at^2 2 = 1/2 * 9.8t^2 2 = 4.9t^2 t = 0.64 s The object takes 0.64 seconds to hit the ground. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:21:04 `q002. While an object dropped from rest falls 2 meters under the influence of gravity, another object moves along a level surface at 12 meters/second. How far does the second object move during the time required for the first object to fall?
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RESPONSE --> OK confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:25:15 `q003. An object rolls off the edge of a tabletop and falls to the floor. At the instant it leaves the edge of the table is moving at 6 meters/second, and the distance from the tabletop to the floor is 1.5 meters. Since if we neglect air resistance there is zero net force in the horizontal direction, the horizontal velocity of the object will remain unchanged. Since the gravitational force acts in the vertical direction, motion in the vertical direction will undergo the acceleration of gravity. Since at the instant the object leaves the tabletop its motion is entirely in the horizontal direction, the vertical motion will also be characterized by an initial velocity of zero. How far will the object therefore travel in the horizontal direction before it strikes the floor?
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RESPONSE --> VERTICAL v0 = 0m/s a = 9.8m/s^2 d = 1.5m 1.5 = 1/2 9.8t^2 t = 0.55s HORIZONTAL v = 6m/s t = 0.55s v = d/t 6 = d/.55 d = 3.3 m The object will travel 3.3 meters in the horizontal direction before it strikes the floor. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:25:30 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 1.5 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 1.5 m / (9.8 m/s^2) ) = .54 sec, approx., so the object falls for about .54 seconds. The horizontal motion will therefore last .54 seconds. Since the initial 6 meter/second velocity is in the horizontal direction, and since the horizontal velocity is unchanging, the object will travel `ds = 6 m/s * .54 s = 3.2 m, approximately.
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RESPONSE --> OK self critique rating: 3
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17:29:35 `q004. An object whose initial velocity is in the horizontal direction descends through a distance of 4 meters before it strikes the ground. It travels 32 meters in the horizontal direction during this time. What was its initial horizontal velocity? What are its final horizontal and vertical velocities?
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RESPONSE --> VERTICAL v0 = 0m/s d = 4m a = 9.8 vf^2 = v0^2 + 2ad vf^2 = 0 + 2*9.8*4 = 78.4 vf = 8.85 m/s 4 = 1/2 9.8t^2 t^2 = .816 t = 0.9s HORIZONTAL d = 32m t = 0.9s v = d/t v = 35.56m/s The final vertical velocity is 8.85m/s, the final and intial horizontal velocities are both 35.56m/s since it isn't accelerating in the horizontal direction. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:29:57 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 4 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 4 m / (9.8 m/s^2) ) = .9 sec, approx., so the object falls for about .9 seconds. The horizontal displacement during this .9 second fall is 32 meters, so the average horizontal velocity is 32 meters/(.9 second) = 35 meters/second, approximately. The final vertical velocity is easily calculated. The vertical velocity changes at a rate of 9.8 meters/second^2 for approximately .9 seconds, so the change in vertical velocity is `dv = 9.8 m/s^2 * .9 sec = 8.8 m/s. Since the initial vertical velocity was zero, final vertical velocity must be 8.8 meters/second in the downward direction. The final horizontal velocity is 35 meters/second, since the horizontal velocity remains unchanging until impact.
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RESPONSE --> OK self critique rating: 3
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