course Phy 201 ꗄLq֧Jassignment #021
......!!!!!!!!...................................
19:48:13 `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?
......!!!!!!!!...................................
RESPONSE --> Since we know the horizontal velocity will stay constant we need to find the vertical velocity. Since we know `ds,a and v0 we find vf by using the equation vf^2=v0^2+2a`ds. We fill in the equation vf=sqrt(2(9.8m/s^2)(3m),vf=7.7m/s.Now ew know the x component and y component.x=12m/s and y=7.7m/s, but since y is the vertical down component it will be -7.7m/s. We can know find the the magntude byy using the pythag. theorem. magnitude=sqrt(12m/s^2+-7.7m/s^2), magnitude=14.3m/s. We can now find the direction of the vector using thhw tan^-1(y/x), tan^-1(-7.7m/s/12m/s)=-33degrees. If we add -33deg to 360deg we get 327deg confidence assessment: 2
.................................................
......!!!!!!!!...................................
19:48:52 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:20:11 `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?
......!!!!!!!!...................................
RESPONSE --> Since the angle is pointing upward in the first quadrant of the graph we make the y direction positive and the intial vertical velocity. We find the y component by taking intial velocity 20m/s*sin(30deg)=10m/s.We now know v0=10m/s,a=9/8m/s^2,and`ds=12m. We can now use vf^2=v0^2+2a`ds to find vf an dthen solve for `dt.vf=sqrt((10m/s)^2+2*9.8m/s^2*12m),vf=sqrt(335.2),vf=18.3m/s.Since the intial velociyt is positive in the upward direction then the vf is in the downward direction so it will be -18.3m/s. We know `dt=`ds/vAve we find vAve by taking vAve=(10m/s+-18.3m/s)/2,vAve=-4.2m/s, `dt=12m/-4.2m/s,`dt=-2.9s,We cannot have a -time something is wrong here confidence assessment: 1
.................................................
......!!!!!!!!...................................
21:24:03 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.
......!!!!!!!!...................................
RESPONSE --> I did not make my displacement -12meters. i guess sine the upward direction is positive and the level surface is below the displacement will be negitive self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:28:29 `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.
......!!!!!!!!...................................
RESPONSE --> To find the hprizontal velocity we do the same as we did for y. We take 20m/s*cos(30deg)=17.3m/s Since the horizontal velocity will remain constant then at 17.3m/s for the time period of 2.9s we get `ds=vAve*`dt,`ds=17.3m/s*2.9s,`ds=50.2m confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:29:36 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
"