course phy 121 ^Ӝ̨assignment #016 016. Projectiles Physics II 08-26-2008
......!!!!!!!!...................................
22:02:06 `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?
......!!!!!!!!...................................
RESPONSE --> vf^2 = v0^2 + 2a*ds vf^2 = 0 + 2(9.8m/s^2) * 2m vf^2 = 39.2 vf = 6.26m/s vAve = (vf+v0)/2 vAve = (6.26m/s + 0)/2 = 3.13m/s dt =ds/vAve dt = 2m / 3.13m/s = .63s confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:05:14 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.
......!!!!!!!!...................................
RESPONSE --> I chose the hard way to do it. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:07:48 `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?
......!!!!!!!!...................................
RESPONSE --> ds = vAve * dt ds = 12m/s * .64s = 7.68m confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:08:04 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:21:47 `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?
......!!!!!!!!...................................
RESPONSE --> horiz vel = 6 m/s horiz vel doesn't change vert ds = 1.5m vert v0 = 0 a = 9.8m/s^2 what is horiz ds? horiz ds = horiz vAve * dt Have to find dt first vert vf^2 = vert v0^2 + 2a* vert ds vf^2 = 0 + 2 (9.8m/s^2) * 1.5m vf^2 = 29.4 vert vf = 5.4m/s dt = ds/vAve vAve = 5.4m/s / 2 = 2.7m/s dt = 1.5m / 2.7m/s = 4.05s Now plug that into first equation horiz ds = horiz vAve * dt = 6 m/s * 4.05s horiz ds = 24.3 m confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:32:36 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.
......!!!!!!!!...................................
RESPONSE --> My way would have worked if I had used vf = v0 + a * dt to find dt instead of dt = ds/vAve or if I had actually divided ds by vAve instead of making a silly math error and multiplying the two numbers! The procedure you demonstrate would have been less steps and easier though. I think self critique assessment: 2
.................................................
......!!!!!!!!...................................
23:01:37 `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?
......!!!!!!!!...................................
RESPONSE --> vert ds = 4m horiz ds = 32m a = 9.8m/s^2 horiz vel is unknown but we will assume it is unchanged from initial to final just like the last problem. vert vel is unknown but we will assume it starts at zero just like the last problem Solve vertical first ds = v0*dt + 1/2 a * dt^2 4m = 0 + 1/2 (9.8m/s^2) * dt^2 4m / 4.9 m/s^2 = dt^2 .816 = dt^2 dt = .9s vert vf = vert v0 + a * dt vert vf = 0 +9.8m/s^2 * .9s = 8.82m/s So now I have time and horiz distance vAve = ds/dt vAve = 32m / .9s = 35.6m/s Since we are assuming not horiz accel that means that there is no change in horz vel so vf and v0 are both 35.6m/s vAVe = ds/dt vAve = 32m/.9s = 35.6m/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
23:02:43 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.
......!!!!!!!!...................................
RESPONSE --> Yay! I can't believe I did it right! self critique assessment: 3
.................................................