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D 6. This equation yields two solutions: t = 3.96 and t = –1.03. [latex]R=\frac{{{v}_{0}}^{2}\sin{2\theta }_{0}}{g}\\[/latex]. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. 57.14 N, 5.) Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) Calculate the time it takes the rock to follow this path. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation. 26. Because gravity is vertical, ax=0. This unit is the study motion, in other words this unit is a field of science that describes how an object or system creates forces that causes motion. dx=-5.12 m/s, dy=-2.72 m/s, 4.) It is important to set up a coordinate system when analyzing projectile motion. In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0º above the horizontal? 9. The kinematic equations for horizontal and vertical motion take the following forms: Step 3. derive the relevant equations of motion along each direction. Set the angle, initial speed, and mass. Match. Learn. What distance does the ball travel horizontally? Initial values are denoted with a subscript 0, as usual. Part 1: 1.0-1.4 The first unit we looked at in our journey through physics 20 is Kinematics. When would it be necessary for the archer to use the larger angle? In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height. Its solutions are given by the quadratic formula: [latex]t=\frac{-bpm \sqrt{{b}^{2}-4\text{ac}}}{\text{2}\text{a}}\\[/latex]. HW 3.1 Answers: 1.) UNIT 1 Kinematics NAME Scenario 1 .N Projectile Motion Part 2. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed of 40.0 m/s (144 km/h). D 20. The range R of a projectile on level ground for which air resistance is negligible is given by. He used it to predict the range of a projectile. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). In each case shown here, a projectile is launched from a very high tower to avoid air resistance. It is given by v0y = v0 sin θ, where v0y is the initial velocity of 70.0 m/s, and θ0 = 75.0º is the initial angle. Blast a Buick out of a cannon! However, to simplify the notation, we will simply represent the component vectors as x and y.). the motion of a projectile such that the horizontal component of the velocity is constant, and the vertical motion has a constant acceleration due to gravity. 19. 149.2 kW, 4.) (b) What maximum height does it reach? 13. θ =6.1º. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. Figure 1 illustrates the notation for displacement, where s is defined to be the total displacement and x and y are its components along the horizontal and vertical axes, respectively. [latex]y=\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex]. How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. (a) The greater the initial speed v0, the greater the range for a given initial angle. (b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. The trajectory of a fireworks shell. On the same set of axes in Part A, sketch the graphs, for the seconG rocN. This time is also reasonable for large fireworks. Because air resistance is negligible, ax=0 and the horizontal velocity is constant, as discussed above. The initial angle θ0 also has a dramatic effect on the range, as illustrated in Figure 5(b). Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? (b) What must have been the initial horizontal component of the velocity? If air resistance is considered, the maximum angle is approximately 38º. Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? ____B.____ A car travels 30 kilometers at an average speed of 60 kilometers per hour and then 30 kilometers at an average speed of 30 kilometers per hour. Its magnitude is s, and it makes an angle θ with the horizontal. The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. Construct Your Own Problem Consider a ball tossed over a fence. ,f the JraShs for 3art % are GiƬerent than, those for 3art $± use GiƬerent colors or GiƬerent lines ²e.J.±, GasheG Ys. Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. describe the projectile motion in two separate directions, vertical and horizontal directions, and emphasise the essential facts about the motion along each direction. Will the ball land in the service box, whose out line is 6.40 m from the net? You should obtain an equation of the form [latex]y=\text{ax}+{\text{bx}}^{2}\\[/latex] where a and b are constants. 1 March 18, 2015 Unit 1 : Kinematics. The time a projectile is in the air is governed by its vertical motion alone. During a lecture demonstration, a professor places two coins on the edge of a table. (See Figure 4.). Both accelerations are constant, so the kinematic equations can be used. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º nor 90º): (a) Is the acceleration ever zero? She then flicks one of the coins horizontally off the table, simultaneously nudging the other over the edge. This is called escape velocity. A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. A projectile is launched from the ground at an angle of 30 degrees. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. Add air resistance. Learn about projectile motion by firing various objects. (b) When is the velocity a minimum? A 16. Thus, vOy = v0 sin θ0 = (70.0 m/s)(sin 75º) = 67.6 m/s. (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground? (These equations describe the x and y positions of a projectile that starts at the origin.) - 0.531 '1'1. Projectile motion (part 1) Projectile motion (part 2) This is the currently selected item. The magnitudes of these vectors are s, x, and y. Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The object thus falls continuously but never hits the surface. 4.23 m. No, the owl is not lucky; he misses the nest. In this case, the easiest method is to use [latex]y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex]. Base your answers to questions 27 through 29 on the information and diagram below. LO: 3.A.1.2 The student is able to design an experimental investigation of the motion of an object. The time for projectile motion is completely determined by the vertical motion. D 17. (c) The ocean is not flat, because the Earth is curved. [latex]x-{x}_{0}={v}_{0x}t=\left({v}_{0}\cos\theta \right)t=R\\[/latex], and substituting for t gives: [latex]R={v}_{0}\cos\theta \left(\frac{{2v}_{0}\sin\theta}{g}\right)=\frac{{{2v}_{0}}^{2}\sin\theta \cos\theta }{g}\\[/latex]. 25. Uniform Circular Motion Equation for Period. C 18. 9. Such descriptions can rely upon words, diagrams, graphics, numerical data, and mathematical equations. Verify the ranges shown for the projectiles in Figure 5(b) for an initial velocity of 50 m/s at the given initial angles. PLAY. (a) 24.2 m/s (b) The ball travels a total of 57.4 m with the brief gust of wind. Prove that the trajectory of a projectile is parabolic, having the form [latex]y=\text{ax}+{\text{bx}}^{2}\\[/latex]. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) A projectile is launched over level ground at 85 m/s, 25 ° above the horizontal. 17. D 13. Apply the principle of independence of motion to solve projectile motion problems. Why does the punter in a football game use the higher trajectory? In Addition of Velocities, we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic. A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. (a) 560 m/s (b) 800 × 103 m (c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b). Substituting known values yields. Is the owl lucky enough to have the mouse hit the nest? A projectile is launched into the air with an initial speed of v i at a launch angle of 30° above the horizontal. Unit 3 Newton's Laws of Motion. [latex]y=\frac{\left(67.6\text{ m/s}\right)^{2}}{2\left(9.80\text{ m/s}^{2}\right)}\\[/latex], Kinematics in Two Dimensions: An Introduction, Vector Addition and Subtraction: Analytical Methods, http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics. By “height” we mean the altitude or vertical position y above the starting point. An arrow is shot from a height of 1.5 m toward a cliff of height H. It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. 24. A 2. The correct answer is d. The rock is accelerating constantly at 10 m/s2, so its displacement can be calculated using simple kinematics: € Δy=v i t+1 2 at2 Δy=0+1 2 (−10m/s2)(7s)2 Δy=245m It is arguably easier to calculate this quickly by determining the average velocity during the seven seconds of The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described. (a) 18.4º (b) The arrow will go over the branch. To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m. 18. Determine the location and velocity of a projectile at different points in its trajectory. An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Recombine the horizontal and vertical components of location and/or velocity using the following equations: 1. Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. Derive [latex]R=\frac{{{v}_{0}}^{2}\text{\sin}{2\theta }_{0}}{g}\\[/latex] for the range of a projectile on level ground by finding the time t at which y becomes zero and substituting this value of t into the expression for x – x0, noting that R = x – x0. [latex]\bar{v}=\frac{{v}_{0}+v}{2}\\[/latex], [latex]x={x}_{0}+{v}_{0}t+\frac{1}{2}{\text{at}}^{2}\\[/latex]. Yes, the ball lands at 5.3 m from the net. (c) Is the acceleration ever opposite in direction to a component of velocity? Resolve or break the motion into horizontal and vertical components along the x- and y-axes. Call the maximum height y=h; then. It lands on the top edge of the cliff 4.0 s later. 3.   Terms. 8. 16. where v0y was found in part (a) to be 14.3 m/s. The components of position s are given by the quantities x and y, and the components of the velocity v are given by vx = v cos θ and vy = v sin θ, where v is the magnitude of the velocity and θ is its direction. In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon. This is true only for conditions neglecting air resistance. Gravity. 2-Dimensional projectile motion can be broken down into 2 1-dimensional motion problems. Explain your answer. (d) The x – and y -motions are recombined to give the total velocity at any given point on the trajectory. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally. D Part 2 – Diagrams Communication – 12 marks 1. Kilauea in Hawaii is the world’s most continuously active volcano. At what angle above the horizontal must the ball be thrown to exactly hit the basket? soliG³ to GiƬerentiate between the lines in 3art $, anG 3art %. (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket? (4 marks) This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity. Unreasonable Results (a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s. With a large enough initial speed, orbit is achieved. (c)What maximum height is attained by the ball? The horizontal displacement is horizontal velocity multiplied by time as given by x = x0 + vxt, where x0 is equal to zero: where vx is the x-component of the velocity, which is given by vx = v0 cos θ0 Now, vx = v0 cos θ0 = (70.0 m/s)(cos 75º) = 18.1 m/s, The time t for both motions is the same, and so x is. We can find the time for this by using. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. [latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\\[/latex]. (See Figure 6.) E 12. The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. 19. axes provided below the diagram to the right: in Part A. vx=3.27 m/s, vy=2.29 m/s, 4.) For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with θ0 = 45º. |25 UNIT 1 Kinematics | 1.N Projectile Motion Part 2. It strikes a target above the ground 3.00 seconds later. (a) −0.486 m (b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. 1. Now we must find v0y, the component of the initial velocity in the y-direction. When we speak of the range of a projectile on level ground, we assume that R is very small compared with the circumference of the Earth. ... Unit 1 Kinematics. State your assumptions. *This is the entire unit for two-dimensional kinematics or projectile motion. 4.A.2.3 The stu- The final vertical velocity is given by the following equation: [latex]{v}_{y}={v}_{0y}\text{gt}\\[/latex]. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. [latex]y-{y}_{0}=0={v}_{0y}t-\frac{1}{2}{gt}^{2}=\left({v}_{0}\sin\theta\right)t-\frac{1}{2}{gt}^{2}\\[/latex] , so that [latex]t=\frac{2\left({v}_{0}\sin\theta \right)}{g}\\[/latex]. 0.315 hp, 8.) An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. This example asks for the final velocity. Kinematics equation without v. Deltax=vot+ (1/2)at^2. Thus. vx=1.11 m/s, vy=-3.42 m/s, 3.) B 15. Ball 2 has a larger horizontal velocity than Ball 1 C. Ball 2 has less mass than Ball 1 D. It is not possible for Ball 2 to travel further than Ball 1 The 2 balls shown were launched at the same speed. worked examples on a two-dimension projectile motion. C 11. (Neglect air resistance.). Therefore: vx = v0 cos θ0 = (25.0 m/s)(cos 35º) = 20.5 m/s. Note that the final vertical velocity, vy, at the highest point is zero. [latex]y=\frac{\left(67.6\text{ m/s}\right)^{2}}{2\left(9.80\text{ m/s}^{2}\right)}\\[/latex]. A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. A player standing on the free throw line throws the ball with an initial speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. SPH3U1 September 28, 2017 Unit 1 Test – Kinematics ANSWERS Part 1 – Multiple Choice Knowledge and Understanding – 20 marks 1. 4. The projectile lands on the ground 2.0 seconds later. (a) What is the height of the cliff? (a) At what speed does the ball hit the ground? (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes.   Privacy In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? where v0 is the initial speed and θ0 is the initial angle relative to the horizontal. Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. (d) Can the speed ever be the same as the initial speed at a time other than at t = 0? ... horizontal components of the motion. (a) What vertical velocity does he need to rise 0.750 m above the floor? 302 m/s @ 6.65°, 3.) 67.62 s, 7.) To solve projectile motion problems, perform the following steps: The maximum horizontal distance traveled by a projectile is called the. T= (2pi*r)/v. A football player punts the ball at a 45º angle. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. (c) Can the velocity ever be the same as the initial velocity at a time other than at t = 0? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s. As is customary, we call the horizontal axis the x-axis and the vertical axis the y-axis. If we continued this format, we would call displacement s with components sx and sy. 25. [latex]R=\frac{{{{v}_{0}}}^{}}{\sin{2\theta }_{0}g}\\[/latex], For θ = 45º, [latex]R=\frac{{{{v}_{0}}}^{2}}{g}\\[/latex], R = 91.9 m for v0 = 30 m/s; R = 163 m for v0; R = 255 m for v0 = 50 m/s. Suppose a soccer player kicks the ball from a distance 30 m toward the goal. 0.408 m/s, 2450 J, 9.) D 4. Assume that the radius of the Earth is 6.37 × 103. 673.29 N, 2.) (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (a) Calculate the initial velocity of the shell. Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory. NAME DATE Scenario A rock is thrown horizontally with speed v from the top of a cliƬ oI height H as shown in the diagram to the right. What are the x and y distances from where the projectile was launched to where it lands? Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. 3. [latex]y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex], [latex]{v}_{y}={v}_{0y}-\text{gt}\\[/latex], [latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\mathrm{gt}}^{2}\\[/latex]. Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. Construct a problem in which you calculate the ball’s needed initial velocity to just clear the fence. (b) The effect of initial angle θ0 on the range of a projectile with a given initial speed. (c) How long did this pass take? 22. aisling99. Unit 1 Kinematics 22. (b) How long does it take to get to the receiver? Download File. [latex]h=\frac{{{v}_{0y}}^{2}}{2g}\\[/latex]. Ball 2 accelerates downwards towards the ground slower B. 11. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. unit 1 - kinematics packet answer key - testa.pdf Unit 1 - Kinematics Packet ANSWER KEY - Testa.pdf 4.48 MB (Last Modified on September 18, 2019) Comments (-1) (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. Next class: Kinematics and Vectors Quiz ; Bring Computer to class next day!! Decide who is your partner (2 per group). Question 2 in a three part physics series on projectile motion and 2 dimensional kinematics. This possibility was recognized centuries before it could be accomplished. Verify the ranges for the projectiles in Figure 5 (a) for θ = 45º and the given initial velocities. (c) What is the vertical component of the velocity just before the ball hits the ground? 7. The magnitudes of the components of the velocity v are Vx = V cos θ and Vy = v sin θ where v is the magnitude of the velocity and θ is its direction, as shown in 2. 2205 W a=v^2/r. (c) Is the premise unreasonable or is the available equation inapplicable? E 19. Documents and powerpoints for this unit are here below. [latex]s=\sqrt{{x}^{2}+{y}^{2}}\\[/latex], [latex]v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}\\[/latex]. How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (c) What is the arrow’s impact speed just before hitting the cliff? &reate a Ne\ so that it is eas\ to GiƬerentiate the. (Another way of finding the time is by using [latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\\[/latex], and solving the quadratic equation for t.).

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