Spider-Man stands at the edge of a tall building. He pauses for a moment to calculate how to get to another building of the same height, 70.0 m away. After calculating, he jumps straight up, with negligible horizontal speed, but enough to just miss the edge of the building he was on. He hen falls 20.0 m below his initial elevation, at which point he becomes horizontally aligned with a par. He shoots his web at this bar (assume it travels so fast that it latches onto the bar almost nstantaneously), and uses it to swing in a circular arc of radius 30.0 m. When his web makes an angle of 15.0° below the horizontal on the opposite side of the bar, he lets go of it and moves in ree-fall again, just clearing the top of the opposite building. His trajectory is shown in Figure 2. 20.0 m 40.0 m 15.0° 30.0 m 3 Figure 2: Spider-Man's trajectory is shown in red. (a) What speed must Spider-Man have at point 2 when he lets go of the web, in order to just clear the top of the opposite building? (b) With what speed must Spider-Man jump upward at point 1 to achieve the desired speed at point 2? (c) Given this trajectory, when Spider-Man is at the bottom of his swing (point 3), how many times his weight will the tension in his web be? 2.
Spider-Man stands at the edge of a tall building. He pauses for a moment to calculate how to get to another building of the same height, 70.0 m away. After calculating, he jumps straight up, with negligible horizontal speed, but enough to just miss the edge of the building he was on. He hen falls 20.0 m below his initial elevation, at which point he becomes horizontally aligned with a par. He shoots his web at this bar (assume it travels so fast that it latches onto the bar almost nstantaneously), and uses it to swing in a circular arc of radius 30.0 m. When his web makes an angle of 15.0° below the horizontal on the opposite side of the bar, he lets go of it and moves in ree-fall again, just clearing the top of the opposite building. His trajectory is shown in Figure 2. 20.0 m 40.0 m 15.0° 30.0 m 3 Figure 2: Spider-Man's trajectory is shown in red. (a) What speed must Spider-Man have at point 2 when he lets go of the web, in order to just clear the top of the opposite building? (b) With what speed must Spider-Man jump upward at point 1 to achieve the desired speed at point 2? (c) Given this trajectory, when Spider-Man is at the bottom of his swing (point 3), how many times his weight will the tension in his web be? 2.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:Spider-Man stands at the edge of a tall building. He pauses for a moment to calculate how to
get to another building of the same height, 70.0 m away. After calculating, he jumps straight up,
with negligible horizontal speed, but enough to just miss the edge of the building he was on. He
then falls 20.0 m below his initial elevation, at which point he becomes horizontally aligned with a
bar. He shoots his web at this bar (assume it travels so fast that it latches onto the bar almost
instantaneously), and uses it to swing in a circular arc of radius 30.0m. When his web makes an
angle of 15.0° below the horizontal on the opposite side of the bar, he lets go of it and moves in
free-fall again, just clearing the top of the opposite building. His trajectory is shown in Figure 2.
1
20.0 m
40.0 m
15.0°
30.0 m
2
3
Figure 2: Spider-Man's trajectory is shown in red.
(a) What speed must Spider-Man have at point 2 when he lets go of the web, in order to just clear
the top of the opposite building?
(b) With what speed must Spider-Man jump upward at point 1 to achieve the desired speed at
point 2?
(c) Given this trajectory, when Spider-Man is at the bottom of his swing (point 3), how many times
his weight will the tension in his web be?
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