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Having gained entry into a local aquarium, a bobcat launches himself into projectile motion with a velocity of 12.0 m/s directed 30 degrees above the horizontal. At the instant of his launch, the cat is located 15.0 m above the ground, at a horizontal distance of 12.0 m from a large fish tank, and 6.00 m above the water level in the tank, as shown in the figure below. In order to land in the fish tank, the cat must make it over the top of the near wall of the fish tank, which is located 3.00 m below the level of the cat's launch point. Of course, the cat does not want to "overshoot" the entire 10.0 m wide tank and land on the hard ground on the other side! Treat the cat as a point particle (or a ball which has been thrown). Assume that air resistance is negligible. Assume that g=9.80m/2. NEAR WALL OF THE FISH TANK 30 13.00 m +x 6.00 m 1. 15.0 m 1. a bobcat 3. 12.0 m 10.0 m THE GROUND. 1. How much time does it take the cat to reach the highest point in his trajectory? A. 1.22 s. В. 1.06 s. С. 1.25 s. D. 0.612 s. E. 0.817 s. 2. When the cat passes through the highest point in his trajectory, A. his velocity is zero. B. his acceleration is equal to zero. C the y-component of his acceleration is equal to zero. D. his speed is equal to zero. E. None of the above. 3. How much time would it take the cat to travel a horizontal distance of 12.0 m from his launch point? C. 1.00 s. A. 2.00 s. В. 1.15 s. D. 1.22 s. E. 1.73 s. 4. At the instant when the cat has traveled a horizontal distance of 12.0 m from his launch point, the scalar y component of the cat's velocity would be

Having gained entry into a local aquarium, a bobcat launches himself into projectile motion with a velocity of 12.0 m/s directed 30 degrees above the horizontal. At the instant of his launch, the cat is located 15.0 m above the ground, at a horizontal distance of 12.0 m from a large fish tank, and 6.00 m above the water level in the tank, as shown in the figure below. In order to land in the fish tank, the cat must make it over the top of the near wall of the fish tank, which is located 3.00 m below the level of the cat's launch point. Of course, the cat does not want to "overshoot" the entire 10.0 m wide tank and land on the hard ground on the other side! Treat the cat as a point particle (or a ball which has been thrown). Assume that air resistance is negligible. Assume that g=9.80m/2. NEAR WALL OF THE FISH TANK 30 13.00 m +x 6.00 m 1. 15.0 m 1. a bobcat 3. 12.0 m 10.0 m THE GROUND. 1. How much time does it take the cat to reach the highest point in his trajectory? A. 1.22 s. В. 1.06 s. С. 1.25 s. D. 0.612 s. E. 0.817 s. 2. When the cat passes through the highest point in his trajectory, A. his velocity is zero. B. his acceleration is equal to zero. C the y-component of his acceleration is equal to zero. D. his speed is equal to zero. E. None of the above. 3. How much time would it take the cat to travel a horizontal distance of 12.0 m from his launch point? C. 1.00 s. A. 2.00 s. В. 1.15 s. D. 1.22 s. E. 1.73 s. 4. At the instant when the cat has traveled a horizontal distance of 12.0 m from his launch point, the scalar y component of the cat's velocity would be

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter3: Motion Along A Straight Line
Section: Chapter Questions
Problem 64P: An express train passes through a station. It enters with an initial velocity of 22.0 m/s and...
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Having gained entry into a local aquarium, a bobcat launches himself into projectile motion with a velocity of 12.0 m/s directed 30 degrees above the horizontal. At the instant of his launch, the cat is located 15.0 m above the ground, at a horizontal distance of 12.0 m from a large fish tank, and 6.00 m above the water level in the tank, as shown in the figure below. In order to land in the fish tank, the cat must make it over the top of the near wall of the fish tank, which is located 3.00 m below the level of the cat's launch point. Of course, the cat does not want to "overshoot" the entire 10.0 m wide tank and land on the hard ground on the other side! Treat the cat as a point particle (or a ball which has been thrown). Assume that air resistance is negligible. Assume that g=9.80"/ 2 Vi NEAR WALL OF THE FISH TANK +y 130° 13.00 m +x 6.00 m 15.0 m a bobcat 12.0 m 10.0 m THE GROUND 1. How much time does it take the cat to reach the highest point in his trajectory? A. 1.22 s. B. 1.06 s. C. 1.25 s. D. 0.612 s. E. 0.817 s. 2. When the cat passes through the highest point in his trajectory, A. his velocity is zero. B. his acceleration is equal to zero. C the y-component of his acceleration is equal to zero. D. his speed is equal to zero. E. None of the above. 3. How much time would it take the cat to travel a horizontal distance of 12.0 m from his launch point? C. 1.00 s. A. 2.00 s. B. 1.15 s. D. 1.22 s. E. 1.73 s. 4. At the instant when the cat has traveled a horizontal distance of 12.0 m from his launch point, the scalar y component of the cat's velocity would be A. 13.6 m/s. B. - 3.62 m/s. C. -5.96 m/s D. 3.80 m/s. E. -5.32 m/s.
Transcribed Image Text:Having gained entry into a local aquarium, a bobcat launches himself into projectile motion with a velocity of 12.0 m/s directed 30 degrees above the horizontal. At the instant of his launch, the cat is located 15.0 m above the ground, at a horizontal distance of 12.0 m from a large fish tank, and 6.00 m above the water level in the tank, as shown in the figure below. In order to land in the fish tank, the cat must make it over the top of the near wall of the fish tank, which is located 3.00 m below the level of the cat's launch point. Of course, the cat does not want to "overshoot" the entire 10.0 m wide tank and land on the hard ground on the other side! Treat the cat as a point particle (or a ball which has been thrown). Assume that air resistance is negligible. Assume that g=9.80"/ 2 Vi NEAR WALL OF THE FISH TANK +y 130° 13.00 m +x 6.00 m 15.0 m a bobcat 12.0 m 10.0 m THE GROUND 1. How much time does it take the cat to reach the highest point in his trajectory? A. 1.22 s. B. 1.06 s. C. 1.25 s. D. 0.612 s. E. 0.817 s. 2. When the cat passes through the highest point in his trajectory, A. his velocity is zero. B. his acceleration is equal to zero. C the y-component of his acceleration is equal to zero. D. his speed is equal to zero. E. None of the above. 3. How much time would it take the cat to travel a horizontal distance of 12.0 m from his launch point? C. 1.00 s. A. 2.00 s. B. 1.15 s. D. 1.22 s. E. 1.73 s. 4. At the instant when the cat has traveled a horizontal distance of 12.0 m from his launch point, the scalar y component of the cat's velocity would be A. 13.6 m/s. B. - 3.62 m/s. C. -5.96 m/s D. 3.80 m/s. E. -5.32 m/s.
a bobcat 12.0 m 1. How much time does it take the cat to reach the highest point in his trajectory? D. 0.612 s. A. 1.22 s. B. 1.06 s. C. 1.25 s. E. 0.817 s. 2. When the cat passes through the highest point in his trajectory, A. his velocity is zero. C the y-component of his acceleration is equal to zero. E. None of the above. B. his acceleration is equal to zero. D. his speed is equal to zero. 3. How much time would it take the cat to travel a horizontal distance of 12.0 m from his launch point? C. 1.00 s. А. 2.00 s. B. 1.15 s. D. 1.22 s. E. 1.73 s. 4. At the instant when the cat has traveled a horizontal distance of 12.0 m frem his launch point, the scalar y component of the cat's velocity would be from A. 13.6 m/s. B. - 3.62 m/s. C. -5.96 m/s. D. 3.80 m/s. E. - 5.32 m/s. 5. At the instant when the cat has traveled a horizontal distance of 12.0 m from his launch point, his speed would be A. 15.7 m/s. B. 11.7 m/s. C. 24.0 m/s. D. 11.0 m/s. E. 12.0 m/s. 6. The cat would pass over the near wall of fish tank * 0.39 m above the top of the near wall. ~ 1.1 m above the top of the near wall. - 3.4 m above the top of the near wall. ~ 4.1 m above the top of the near wall. A. В. C. D. E. The cat will hit the ground before reaching the fish tank. 7. True or False: The cat would hit the surface of the water in the tank at a horizontal distance of 7.5 m beyond the near wall of the tank? A. True. B. False.
Transcribed Image Text:a bobcat 12.0 m 1. How much time does it take the cat to reach the highest point in his trajectory? D. 0.612 s. A. 1.22 s. B. 1.06 s. C. 1.25 s. E. 0.817 s. 2. When the cat passes through the highest point in his trajectory, A. his velocity is zero. C the y-component of his acceleration is equal to zero. E. None of the above. B. his acceleration is equal to zero. D. his speed is equal to zero. 3. How much time would it take the cat to travel a horizontal distance of 12.0 m from his launch point? C. 1.00 s. А. 2.00 s. B. 1.15 s. D. 1.22 s. E. 1.73 s. 4. At the instant when the cat has traveled a horizontal distance of 12.0 m frem his launch point, the scalar y component of the cat's velocity would be from A. 13.6 m/s. B. - 3.62 m/s. C. -5.96 m/s. D. 3.80 m/s. E. - 5.32 m/s. 5. At the instant when the cat has traveled a horizontal distance of 12.0 m from his launch point, his speed would be A. 15.7 m/s. B. 11.7 m/s. C. 24.0 m/s. D. 11.0 m/s. E. 12.0 m/s. 6. The cat would pass over the near wall of fish tank * 0.39 m above the top of the near wall. ~ 1.1 m above the top of the near wall. - 3.4 m above the top of the near wall. ~ 4.1 m above the top of the near wall. A. В. C. D. E. The cat will hit the ground before reaching the fish tank. 7. True or False: The cat would hit the surface of the water in the tank at a horizontal distance of 7.5 m beyond the near wall of the tank? A. True. B. False.
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