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. Vi NEAR WALL OF THE FISH TANK 30 3.00 m ++ 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, B. his acceleration is equal to zero. A. his velocity is 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.
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. Vi NEAR WALL OF THE FISH TANK 30 3.00 m ++ 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, B. his acceleration is equal to zero. A. his velocity is 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.
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
Problem 3P: A particle initially located at the origin has an acceleration of a=3.00jm/s2 and an initial...
Related questions
Concept explainers
Topic Video
Question
Can you answer first 4 questions
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.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning