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A Fireworks Projectile Explodes High and Away During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in Figure 3.38. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? Strategy Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a, = 0 and a =-g. We can then define xo and yo to be zero and solve for the desired quantities. Solution for (a) By “height" we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v,,= 0. Since we know the initial and final velocities as well as the initial position, we use the following equation to find y: v = v, – 29(y – Yo). 3.45 h= 233 m Vo 6, = 75° x = 125 m Figure 3.38 The trajectory of a fireworks shell. 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. Because y0 and vy are both zero, the equation simplifies to 0 = vổy – 29y. 3.46 Solving for y gives y= 29 3.47 Now we must find voy, the component of the initial velocity in the y-direction. It is given by Voy = vo sin 0, where voy is the initial velocity of 70.0 m/s, and O0 = 75.0° is the initial angle. Thus, VOy = Vo sin 60 =(70.0 m/s)(sin 75°)= 67.6 m/s. 3.48 and y is (67.6 m/s)² y = 2(9.80 m/s³)' 3.49 so that y= 233m. 3.50

A Fireworks Projectile Explodes High and Away During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in Figure 3.38. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? Strategy Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a, = 0 and a =-g. We can then define xo and yo to be zero and solve for the desired quantities. Solution for (a) By “height" we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v,,= 0. Since we know the initial and final velocities as well as the initial position, we use the following equation to find y: v = v, – 29(y – Yo). 3.45 h= 233 m Vo 6, = 75° x = 125 m Figure 3.38 The trajectory of a fireworks shell. 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. Because y0 and vy are both zero, the equation simplifies to 0 = vổy – 29y. 3.46 Solving for y gives y= 29 3.47 Now we must find voy, the component of the initial velocity in the y-direction. It is given by Voy = vo sin 0, where voy is the initial velocity of 70.0 m/s, and O0 = 75.0° is the initial angle. Thus, VOy = Vo sin 60 =(70.0 m/s)(sin 75°)= 67.6 m/s. 3.48 and y is (67.6 m/s)² y = 2(9.80 m/s³)' 3.49 so that y= 233m. 3.50

Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter3: Motion In Two Dimensions
Section: Chapter Questions
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A Fireworks Projectile Explodes High and Away During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in Figure 3.38. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? Strategy Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a, = 0 and a, =- q. We can then define xo and yo to be zero and solve for the desired quantities. Solution for (a) By “height" we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v, = 0. Since we know the initial and final velocities as well as the initial position, we use the following equation to find y: v = v – 29(y – y0). 3.45 h = 233 m Vo = 75° X= 125 m Figure 3.38 The trajectory of a fireworks shell. 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. Because yo and vy are both zero, the equation simplifies to 0 = v, – 2gy. 3.46 Solving for y gives y = 2g 3.47 Now we must find voy, the component of the initial velocity in the y-direction. It is given by Voy = vo sin 0, where voy is the initial velocity of 70.0 m/s, and 0o = 75.0° is the initial angle. Thus, VOy = vo sin 00 =(70.0 m/s)(sin 75°)= 67.6 m/s. 3.48 and y is (67.6 m/s)? y = 2(9.80 m/s²)' 3.49 so that y = 233m. 3.50
Transcribed Image Text:A Fireworks Projectile Explodes High and Away During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0° above the horizontal, as illustrated in Figure 3.38. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes? Strategy Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a, = 0 and a, =- q. We can then define xo and yo to be zero and solve for the desired quantities. Solution for (a) By “height" we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v, = 0. Since we know the initial and final velocities as well as the initial position, we use the following equation to find y: v = v – 29(y – y0). 3.45 h = 233 m Vo = 75° X= 125 m Figure 3.38 The trajectory of a fireworks shell. 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. Because yo and vy are both zero, the equation simplifies to 0 = v, – 2gy. 3.46 Solving for y gives y = 2g 3.47 Now we must find voy, the component of the initial velocity in the y-direction. It is given by Voy = vo sin 0, where voy is the initial velocity of 70.0 m/s, and 0o = 75.0° is the initial angle. Thus, VOy = vo sin 00 =(70.0 m/s)(sin 75°)= 67.6 m/s. 3.48 and y is (67.6 m/s)? y = 2(9.80 m/s²)' 3.49 so that y = 233m. 3.50
3. In the OpenStax College Physics textbook, read Chapter 2 section 2.7 (Falling Objects) and then read Chapter 3 (Two Dimensional Kinematics). Then for Example 3.4 in the "Projectile Motion" section, change the initial speed "of 70.0 m/s" to “of Z m/s"; also, change the angle to "60 degrees above the horizontal." Then write the answers (a), (b), and (c) to the question below. Z=87
Transcribed Image Text:3. In the OpenStax College Physics textbook, read Chapter 2 section 2.7 (Falling Objects) and then read Chapter 3 (Two Dimensional Kinematics). Then for Example 3.4 in the "Projectile Motion" section, change the initial speed "of 70.0 m/s" to “of Z m/s"; also, change the angle to "60 degrees above the horizontal." Then write the answers (a), (b), and (c) to the question below. Z=87
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