course PHY 121 1/4 021. projectiles 2*********************************************
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Given Solution: At the instant of first contact, any force between the ball and the floor will not have had time to affect the motion of the ball. So we can assume uniform acceleration throughout the interval from the initial instant to the instant of first contact. 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. STUDENT QUESTION: Why did the y component of 7.7m/s become negative? INSTRUCTOR RESPONSE: When we solved for the time of fall we assumed the downward direction to be positive. When we put the entire picture on an xy coordinate plane we had to change our choice of positive direction to agree with the orientation of the y axis. STUDENT QUESTION: I’m still confused by this part of the equation vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7 meters/second. I don’t understand where the +- part comes from. INSTRUCTOR RESPONSE: The solution to any equation of the form x^2 = c is x = +- sqrt(c). For example if x^2 = 25, then x = sqrt(25) = 5 is a solution, but so is x = -sqrt(25) = -5. Another way of writing these solutions is 25^(1/2) = 5 and -(25^(1/2)) = - 5. It is similar with any even power. For example x^4 = c has two solutions, x = c^(1/4) and x = - c^(1/4). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:ent: ---- ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Angle (theta) = 30. Horiz velocity = 20m/s. `ds = 12m. a = 9.8m/s^2. y = 20m/s * sin(30) = 10m/s. So now we are still missing `dt and need to use one of the motion equations to find it. vf^2 = v0^2 + 2 a `ds, filling in what we know, vf^2 = (10m/s)^2 + 2(9.8m/s^2)(12m) vf^2 = 100 + 235.2 vf^2 = 335.2, taking square of both sides, vf = 18.3m/s 10+18.3 /2 = 12m/ 14m/s = .86s. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT QUESTIONS: Why did we use v0y = 20 meters/second * sine (30 degrees) = 10 meters/second? Why would there be an initial velocity? I don’t understand why we would use sin instead of tan. INSTRUCTOR RESPONSE: A condition of the problem is that the initial velocity is 20 m/s at angle 30 degrees. The initial velocity is therefore a vector with vertical and horizontal components. • Vertical and horizontal motions have different acceleration conditions and must therefore be analyzed separately and independently. To analyze the vertical motion we must use the vertical component of the initial velocity. To analyze the horizontal motion we must use the horizontal component of the initial velocity. The vector v0 has magnitude 20 m/s and makes angle 30 deg with the positive x axis, so its components are • v0_x = 20 m/s * cos(30 deg) and • v0_y = 20 m/s * sin(30 deg). If you are thinking in terms of triangles, the 20 m/s quantity is the hypotenuse of the triangle and the components are the legs of the triangle, which are found using the sine and cosine. For reference, the first figure below depicts the right triangle defined by the vector; the second includes the components. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I totally messed up the negative aspects of this problem. I’m having trouble visualizing why these are negative quantities.. I understand that the object is moving downward so that could be negative but I rarely remember using gravity in a negative direction. I did the steps right, just with the wrong numbers.
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:ent: "