#$&* course Phy 201 11/1 2 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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). STUDENT QUESTION: Hmm. I see where I made a mistake, but I’m not too sure why I made the mistake. I don’t get why the vertical velocity is negative. I had already established the direction of acceleration due to gravity as the positive direction, so I figured the vertical velocity would be positive as well. ********************************************* Question: is the vertical velocity negative because in relation to an x-y plane, the downward motion of the ball relates to the negative aspect of the y-axis. INSTRUCTOR RESPONSE: As with any vector, the angle of the final velocity vector is to be specified using a standard x-y coordinate system. The downward direction was chosen as positive when finding the vertical velocity. The conclusion of this analysis was that the final velocity is 7.7 m/s downward. The standard x-y coordinate system was not used in this part of the solution, and is not necessary at this point. However to specify the direction of the final velocity vector, we must now place the horizontal and vertical velocities on a standard x-y system, with the x axis to the right and the y axis upward. On this coordinate system the y velocity, being downward, has to be represented by a vector whose vertical coordinate is negative, as in the given solution. STUDENT COMMENT I must have punched something into the calculator incorrectly because I didn’t end up getting -35 for the direction of the final velocity. If I had calculated this correctly and added 360 to it then it would have given me 325 degrees. INSTRUCTOR RESPONSE You got an angle of less than 1 degree, which wouldn't make sense in terms of the quantities you used. You probably had your calculator in radian mode. To get the angle in degrees the calculator has to be in degree mode. Be sure you always sketch the situation. If the result you get from the calculator is inconsistent with your sketch, as would be the case here, you need to double-check your calculations. If the inconsistency involves an angle you calculated, then the first thing to check is the mode. STUDENT COMMENT Ok, I think I understand, but it is still a little blurry. It seems simple once you figure everything out, but getting everything in the correct place is my problem. INSTRUCTOR RESPONSE You can solve the vertical velocity and the horizontal velocity for this interval. The velocity at impact will therefore be a vector with known vertical and horizontal components. STUDENT COMMENT and QUESTION i got -33 degrees not -35 an i'm not sure why or if it matters. also can you explain a little more of why we have to add 360 to this number? i didn't do that so would that make this value incorrect? We generally want the angle of a vector to be between 0 and 360 degrees, if we're working in degrees, or between 0 and 2 pi if we're working in radians. It saves us having to think about things like positive and negative, and from having to add or subtract multiples of 360 degrees. The estimated arctan used in the given solution is only an approximation. If you calculated the actual arctangent and got -35 degrees, chances are very good that your result is correct. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating:3 ********************************************* 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: v0y=20m/s*sin(20)=10m/s vf=sqrt(v0^2+2a'ds) vf=sqrt(10m/s+2(9.8m/s)(12m) vf=-18.3m/s vAve=(10m/s+-18.3)/2=-4.2m/s -12m/-4.2m/s=2.9s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 and to 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. Assuming the horizontal direction to be to the right, the angle of the velocity vector with the positive x axis is 30 degrees, so 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 (the components are the green vectors along the coordinate axis). STUDENT QUESTION Why is the v0 not equal to 0m/s. I thought with ALL projectiles, it is inferred that the initial velocity in the vertical direction is 0 m/s. And what I have gotten from this explanation, what we have previously learned as the altitude to represent the height of the table, now we have to basically look at altitude as representing the horixontal distance…. in terms of the way the vector triangle is twisted in its representation. INSTRUCTOR RESPONSE In the problems we've seen to this point, the initial vertical velocity was zero. The reason is that the analysis is often much easier when the initial vertical velocity is zero. We learn the principles of the analysis using the simplest possible system that illustrates the important features of the process; then we move beyond the simplest cases and apply those principles to slightly more complicated conditions. The projectile in this situation was given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal. An object moving at 30 degrees to horizontal has a nonzero vertical velocity, which we find using the principles of vector components. In this problem the only difference with earlier projectile problems is that you have to include a nonzero vertical velocity when finding the time of fall and the final vertical velocity. STUDENT QUESTION To solve for the problem, I should have taken 20 m/s * sin (30 degrees). This equals 10 m/s. What value do we call this though exactly? Is it the vertical motion? INSTRUCTOR RESPONSE 20 m/s is the initial velocity of the object, generally designated v0. 10 m/s is the vertical component of the initial velocity, or the initial velocity in the vertical direction. This is generally designated v0_y, as it was in the given solution. STUDENT QUESTION Shouldn't the average velocity be about 14.2 m/s (( 10 m/s + 18.3 m/s )/2 ) and not 4.2 m/s, making the time it takes about .85 seconds??? INSTRUCTOR RESPONSE The initial vertical velocity is upward, the final vertical velocity is downward. So the two velocities have different signs. When added and divided by 2, we get an average velocity of magnitude 4.2 m/s. If the projectile started at angle 30 deg below horizontal the average velocity would have magnitude 14.2 m/s and the fall would take only .85 sec. Starting above horizontal, though, the ball first rises to a height greater than the 12 m, which takes time, then falls back down. STUDENT QUESTIONS how can you assume that the initial velocity is upward into the right? is this because it's angle is 30 degrees and if so why couldn't it be tilted in the other direction, i.e. downward and to the left? is the average velocity of -4.2 m/s only the velocity in the vertical direction still? i'm sorry but these problems really throw me off! INSTRUCTOR RESPONSE In any problem we have to specify our coordinate system and the direction chosen for the given vector quantities. All we know is that the initial velocity is at 30 degrees above horizontal. The horizontal direction goes both right and left. So it would have been perfectly valid to assume the initial velocity to be up and to the left. However we have to make one assumption or the other. The choice made in this problem results in fewer negative signs, so is probably the more natural choice. We have implicitly chosen our coordinate system to be the 'standard' x-y system with x to the right and y upward. Had we chosen to have the initial velocity to be upward and to the left the initial velocity would have been at 150 degrees, as measured counterclockwise from the positive x axis. The choice in the given solution was stated in the phrase 'Taking the initial velocity to be upward and to the right', so that our initial direction is into the first quadrant, at angle 30 degrees. The later phrase 'Assuming the horizontal direction to be to the right' reiterates this choice. Starting with the phrase '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', the rest of the given solution proceeds to solve for `dt, using these quantities. So the remainder of the given solution is for the vertical motion. The -4.2 m/s avera ge velocity therefore applies to the vertical motion. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x=20m/s*cos(20)=17.3 m/s 17.3m/s*2.7s=46m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The horizontal velocity of the projectile will not change so if we can find this horizontal velocity, knowing that the projectile travels for 2.7 seconds we can easily find the horizontal range. The horizontal velocity of the projectile is simply the x component of the velocity: horizontal velocity = 20 meters/second * cosine (30 degrees) = 17.3 meters/second. Moving at this rate for 2.7 seconds the projectile travels distance 17.3 meters/second * 2.7 seconds = 46 meters, approximately. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q004. A projectile is given an initial velocity of 12 meters / second in a direction 37 degrees above horizontal. It is released from a height of 2 meters and lands on a platform whose height is 3 meters. How much time elapses, and how far does it travel in the horizontal direction between release and landing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y=7m/s*sin(37)=4.2m/s vf=sqrt(4.2m/s^2+2*9.8m/s^2*1m) vf=6.1m/s vAve=(6.1m/s+7m/s)/2=6.6m/s 1m/6.6m/s=.15s x=7m/s *cos(37)=5.6m/s 5.6m/s*.15s=.84m confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ "