course Phy 231 10/21/09 016. Projectiles
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Given Solution: The object has initial velocity 0, acceleration 9.8 meters/second^2, and displacement 2 meters. We can easily solve this problem using the equations of motion. Using the equation `ds = v0 `dt + .5 a `dt^2 we note first that since initial velocity is zero the term v0 `dt will be zero and can be dropped from the equation, leaving `ds = .5 a `dt^2. This equation is then easily solved for `dt, obtaining `dt = `sqrt(2 *`ds / a ) = `sqrt(2 * 2 meters / (9.8 m/s^2) ) = .64 second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (12 m/s) * 0.64 s =~ 7.7 m confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: As we have seen in the preceding problem, the first object requires .64 second to fall. The second object will during this time move a distance of 12 meters/second * .64 second = 8 meters, approximately. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.5 m = .5(9.8m/s^2)(dt^2) dt =~ +- .55 s dt =~ .55s and then (6m/s)(.55 s) = 3.3 m confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4 m = .5(9.8m/s^2)(dt^2) dt =~ +-.9s dt = .9 s vAveHorizontal * .9 s = 32 m vAveH = 35.6 m/s vfVertical^2 = v0^2 + 2(a)(ds) vf = +- sqrt(2(9.8m/s^2)(4 m) vf = 8.9 m/s vfHorizontal = 35.6m/s since acceleration is (supposed to be) 0m/s^2 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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.