Civil Engineering (CE) Exam > Civil Engineering (CE) Questions > 1. Given a Ball, m = 0.2 kg, released from re...
Start Learning for Free
1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.?
Most Upvoted Answer
1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air ...
Introduction
When analyzing the motion of a falling ball subject to air resistance, we need to consider both gravitational and drag forces.
Given Parameters
- Mass of the ball (m): 0.2 kg
- Initial height (y0): 500 m
- Drag coefficient (k): 2 × 10^-4 N.s^2/m^2
1. Speed at Which the Ball Hits the Ground
- Gravitational Force (Fg):
Fg = m * g, where g = 9.81 m/s^2
Fg = 0.2 kg * 9.81 m/s^2 = 1.962 N
- Terminal Velocity (Vt):
At terminal velocity, gravitational force equals drag force:
Fg = FD
1.962 N = k * Vt^2
Vt^2 = 1.962 N / (2 × 10^-4 N.s^2/m^2)
Vt ≈ 99.5 m/s
- Final Speed (Vf):
The final speed of the ball just before hitting the ground can be approximated using energy conservation. The ball converts potential energy to kinetic energy and work done against drag.
Vf ≈ Vt (due to long fall time) ≈ 99.5 m/s
2. Ratio of Speed to Terminal Speed
- Speed Ratio:
To find the ratio of the ball's speed to terminal speed:
Speed ratio = Vf / Vt
Speed ratio ≈ 99.5 m/s / 99.5 m/s = 1
Conclusion
The ball hits the ground at approximately 99.5 m/s, and the ratio of its speed to terminal speed is 1, indicating it has reached terminal velocity just before impact.
When analyzing the motion of a falling ball subject to air resistance, we need to consider both gravitational and drag forces.
Given Parameters
- Mass of the ball (m): 0.2 kg
- Initial height (y0): 500 m
- Drag coefficient (k): 2 × 10^-4 N.s^2/m^2
1. Speed at Which the Ball Hits the Ground
- Gravitational Force (Fg):
Fg = m * g, where g = 9.81 m/s^2
Fg = 0.2 kg * 9.81 m/s^2 = 1.962 N
- Terminal Velocity (Vt):
At terminal velocity, gravitational force equals drag force:
Fg = FD
1.962 N = k * Vt^2
Vt^2 = 1.962 N / (2 × 10^-4 N.s^2/m^2)
Vt ≈ 99.5 m/s
- Final Speed (Vf):
The final speed of the ball just before hitting the ground can be approximated using energy conservation. The ball converts potential energy to kinetic energy and work done against drag.
Vf ≈ Vt (due to long fall time) ≈ 99.5 m/s
2. Ratio of Speed to Terminal Speed
- Speed Ratio:
To find the ratio of the ball's speed to terminal speed:
Speed ratio = Vf / Vt
Speed ratio ≈ 99.5 m/s / 99.5 m/s = 1
Conclusion
The ball hits the ground at approximately 99.5 m/s, and the ratio of its speed to terminal speed is 1, indicating it has reached terminal velocity just before impact.
|
Explore Courses for Civil Engineering (CE) exam
|
|
Top Courses for Civil Engineering (CE)View all
1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.?
Question Description
1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about 1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.?.
1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about 1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.?.
Solutions for 1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.? in English & in Hindi are available as part of our courses for Civil Engineering (CE).
Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free.
Here you can find the meaning of 1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.? defined & explained in the simplest way possible. Besides giving the explanation of
1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.?, a detailed solution for 1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.? has been provided alongside types of 1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.? theory, EduRev gives you an
ample number of questions to practice 1. Given a Ball, m = 0.2 kg, released from rest at y0 = 500 m and air resistance, FD = kV2, where k = 2 × 10-4 N. s2/m2 Units: FD(N), V(m/s) then FIND: (a) Speed at which the ball hits the ground, (b) Ratio of speed to terminal speed.? tests, examples and also practice Civil Engineering (CE) tests.
|
|
Explore Courses for Civil Engineering (CE) exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup