#$&* course Phy 201 4/10/11 around 9:14 016. Projectiles
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Given Solution: The object has initial velocity 0, acceleration 9.8 meters/second^2, and displacement 2 meters. We can easily solve this problem using the equations of motion. Using the equation `ds = v0 `dt + .5 a `dt^2 we note first that since initial velocity is zero the term v0 `dt will be zero and can be dropped from the equation, leaving `ds = .5 a `dt^2. This equation is then easily solved for `dt, obtaining `dt = `sqrt(2 *`ds / a ) = `sqrt(2 * 2 meters / (9.8 m/s^2) ) = .64 second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. While an object dropped from rest falls 2 meters under the influence of gravity, another object moves along a level surface at 12 meters/second. How far does the second object move during the time required for the first object to fall? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `dt= .64sec vo= 0m/s vAve = 12m/s 12m/s= `ds/0.64sec `ds= 7.7m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 equal to the product of its average velocity and the time interval. The second object moves at constant velocity 12 m/s, so its average velocity is 12 m/s. Its displacement is therefore * `ds_horizontal = v_Ave_horizontal * `dt = 12 meters/second * .64 second = 8 meters, approximately. Note that we have denoted the horizontal quantities by adding the subscript 'horizontal', to distinguish them from vertical quantities. The average velocity and displacement here are both in the horizontal direction, so here they are denoted v_ave_horizontal and `ds_horizontal for clarity. It's not always necessary to use the extra subscripts, but when in doubt, it's a good idea. STUDENT COMMENT I didn’t mention anything about v_ave horizontal or ‘ds_horizontal, mainly because I don’t really know what they are. But I got the same answer as you. INSTRUCTOR RESPONSE You had a good solution, but you should understand the notation used here and the reasons for using it. v_ave and `ds respectively stand for average velocity and displacement or change in position v_ave_horizontal stands for the average velocity in the horizontal direction, and `ds_horizontal stands for the displacement in the horizontal direction There is also an average vertical velocity, which might have been used in finding the time of fall, and a vertical displacement of 2 meters downward. These quantities would be designated v_Ave_vertical and `ds_vertical, to distinguish them from the very different horizontal quantities. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: v0= 0m/s `ds= 1.5m fNet = 0N a= 9.8m/s^2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 1.5 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 1.5 m / (9.8 m/s^2) ) = .54 sec, approx., so the object falls for about .54 seconds. The horizontal motion will therefore last .54 seconds. Since the initial 6 meter/second velocity is in the horizontal direction, and since the horizontal velocity is unchanging, its average horizontal velocity is the same as its initial horizontal velocity, and the object will travel thru displacement * `ds_horiz = 6 m/s * .54 s = 3.2 m, approximately. STUDENT QUESTION Well it looks like when Vo=0 then we solve for ‘dt, this is all getting very confusing. INSTRUCTOR RESPONSE When v0 = 0 your equation becomes `ds = .5 a `dt^2. You are given `ds and a, so you can solve this equation for `dt. You get `dt = sqrt( 2 `ds / a). When you substitute the values of `ds and a, and simplify, you get `dt = .54 seconds. The `ds_horiz in the last paragraph is the `ds for horizontal motion. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. An object whose initial velocity is in the horizontal direction descends through a distance of 4 meters before it strikes the ground. It travels 32 meters in the horizontal direction during this time. What was its initial horizontal velocity? What are its final horizontal and vertical velocities? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Summarizing the calculation of `dt * v0_vert = 0, `ds_vert = 4 m and a_vert = 9.8 m/s^2 (`ds and a are both directed downward and have here both been regarded as positive, implying that we have taken the downward direction to be positive) * these values of v0, `ds and a imply `dt = .9 sec. 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. Summarizing the calculation of the horizontal velocity: * horizontal acceleration is zero, so horizontal velocity is constant * in particular v_Ave_horiz = v0_horiz = vf_horiz * `ds_horiz = 32 m, as given * v_Ave_horiz = `ds_horiz / `dt = 32 m / (.9 sec) = 35 m/s. 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. Summary of this calculation: * v0_vert = 0, `dt_vert = .9 s, a_vert = 9.8 m/s^2 * `dv = a * `dt, so in particular `dv_vert = a_vert * `dt_vert = 9.8 m/s^2 * .9 s = 8.8 m/s * vf_vert = v0_vert + `dv = 0 + 8.8 m/s = 8.8 m/s * This could also have been obtained using the fourth equation of uniformly accelerated motion, vf^2 = v0^2 + 2 a `ds, with the vertical quantities. STUDENT QUESTION Ok. But I think I missed the explanation of the initial horizontal velocity. I thought it remained constant throughout the fall of the object, so are the v0_horiz and the vf_horiz both 35m/s?? INSTRUCTOR RESPONSE Horizontal velocity remains constant. Since the horizontal velocity is constant, v0_horiz, v_Ave_horiz and vf_horiz must all be the same. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!