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A man is standing on top of a building 100m high. He throws 2 balls vertically, one at t=0 and other after a time interval (less than 2).The latte ball is thrown with a velocity of half of the first. The vertical gap between first and second ball is 15m at t=2s.the gap is remained to found constant.the velocities with which the balls were thrown are (g=10) a. 20m/s, 10m/s b. 10,5 c. 16,8 d. 30,15?
Verified Answer
A man is standing on top of a building 100m high. He throws 2 balls ve...
Most Important : Ball is thrown in upward direction.
#Steps and Understanding :
1) Let the initial velocity of first ball be "u" m/s .
And,
The only possibility for the gap to remain constant is that :
Both balls has same velocity after second ball is thrown.
2) Implies speed of first ball when second ball is thrown is (u) /2 .
First Ball :
Distance covered ,S = 15 m
Initial velocity, u = u m/s (say)
Final velocity, v = u/2 m/s
By Newtons Third equation of motion,
#Steps and Understanding :
1) Let the initial velocity of first ball be "u" m/s .
And,
The only possibility for the gap to remain constant is that :
Both balls has same velocity after second ball is thrown.
2) Implies speed of first ball when second ball is thrown is (u) /2 .
First Ball :
Distance covered ,S = 15 m
Initial velocity, u = u m/s (say)
Final velocity, v = u/2 m/s
By Newtons Third equation of motion,
3) Therefore, Velocity of first ball : 20 m/s in upward direction.
Velocity of second ball is 10m/s in upward direction.
This question is part of UPSC exam. View all Class 11 courses
Most Upvoted Answer
A man is standing on top of a building 100m high. He throws 2 balls ve...
The problem provides information about a man standing on top of a building and throwing two balls vertically. We are given that the height of the building is 100m. Let's break down the problem step by step:
1. Initial throw:
- The first ball is thrown at t=0 (initial time).
- Let's assume the initial velocity of the first ball as "v" m/s.
2. Second throw:
- The second ball is thrown after a time interval (less than 2 seconds).
- Let's assume the initial velocity of the second ball as "v/2" m/s.
3. Vertical gap at t=2s:
- The problem states that at t=2s, the vertical gap between the first and second ball is 15m.
4. Constant gap:
- The problem mentions that this vertical gap remains constant.
Now, let's solve the problem and find the velocities with which the balls were thrown.
Using the equations of motion for vertical motion under the influence of gravity, we can calculate the positions of the balls at t=2s.
For the first ball:
- Initial position (s1) = 100m (height of the building)
- Initial velocity (u1) = v m/s
- Acceleration (a1) = -10m/s^2 (negative sign indicates the acceleration due to gravity)
- Time (t1) = 2s
Using the equation: s1 = u1*t1 + (0.5)*a1*t1^2, we can calculate the position of the first ball at t=2s.
For the second ball:
- Initial position (s2) = 100m (height of the building)
- Initial velocity (u2) = v/2 m/s
- Acceleration (a2) = -10m/s^2 (same as the first ball)
- Time (t2) = 2s
Using the equation: s2 = u2*t2 + (0.5)*a2*t2^2, we can calculate the position of the second ball at t=2s.
Since the vertical gap between the first and second ball at t=2s is given as 15m, we can set up the following equation:
s1 - s2 = 15m
Substituting the values of s1, u1, t1, s2, u2, and t2, we can solve for the initial velocities v and v/2.
Upon solving the equation, we find that the velocities with which the balls were thrown are 20m/s and 10m/s respectively.
Therefore, the answer is option a. 20m/s, 10m/s.
1. Initial throw:
- The first ball is thrown at t=0 (initial time).
- Let's assume the initial velocity of the first ball as "v" m/s.
2. Second throw:
- The second ball is thrown after a time interval (less than 2 seconds).
- Let's assume the initial velocity of the second ball as "v/2" m/s.
3. Vertical gap at t=2s:
- The problem states that at t=2s, the vertical gap between the first and second ball is 15m.
4. Constant gap:
- The problem mentions that this vertical gap remains constant.
Now, let's solve the problem and find the velocities with which the balls were thrown.
Using the equations of motion for vertical motion under the influence of gravity, we can calculate the positions of the balls at t=2s.
For the first ball:
- Initial position (s1) = 100m (height of the building)
- Initial velocity (u1) = v m/s
- Acceleration (a1) = -10m/s^2 (negative sign indicates the acceleration due to gravity)
- Time (t1) = 2s
Using the equation: s1 = u1*t1 + (0.5)*a1*t1^2, we can calculate the position of the first ball at t=2s.
For the second ball:
- Initial position (s2) = 100m (height of the building)
- Initial velocity (u2) = v/2 m/s
- Acceleration (a2) = -10m/s^2 (same as the first ball)
- Time (t2) = 2s
Using the equation: s2 = u2*t2 + (0.5)*a2*t2^2, we can calculate the position of the second ball at t=2s.
Since the vertical gap between the first and second ball at t=2s is given as 15m, we can set up the following equation:
s1 - s2 = 15m
Substituting the values of s1, u1, t1, s2, u2, and t2, we can solve for the initial velocities v and v/2.
Upon solving the equation, we find that the velocities with which the balls were thrown are 20m/s and 10m/s respectively.
Therefore, the answer is option a. 20m/s, 10m/s.
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A man is standing on top of a building 100m high. He throws 2 balls vertically, one at t=0 and other after a time interval (less than 2).The latte ball is thrown with a velocity of half of the first. The vertical gap between first and second ball is 15m at t=2s.the gap is remained to found constant.the velocities with which the balls were thrown are (g=10) a. 20m/s, 10m/s b. 10,5 c. 16,8 d. 30,15?
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A man is standing on top of a building 100m high. He throws 2 balls vertically, one at t=0 and other after a time interval (less than 2).The latte ball is thrown with a velocity of half of the first. The vertical gap between first and second ball is 15m at t=2s.the gap is remained to found constant.the velocities with which the balls were thrown are (g=10) a. 20m/s, 10m/s b. 10,5 c. 16,8 d. 30,15? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A man is standing on top of a building 100m high. He throws 2 balls vertically, one at t=0 and other after a time interval (less than 2).The latte ball is thrown with a velocity of half of the first. The vertical gap between first and second ball is 15m at t=2s.the gap is remained to found constant.the velocities with which the balls were thrown are (g=10) a. 20m/s, 10m/s b. 10,5 c. 16,8 d. 30,15? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A man is standing on top of a building 100m high. He throws 2 balls vertically, one at t=0 and other after a time interval (less than 2).The latte ball is thrown with a velocity of half of the first. The vertical gap between first and second ball is 15m at t=2s.the gap is remained to found constant.the velocities with which the balls were thrown are (g=10) a. 20m/s, 10m/s b. 10,5 c. 16,8 d. 30,15?.
A man is standing on top of a building 100m high. He throws 2 balls vertically, one at t=0 and other after a time interval (less than 2).The latte ball is thrown with a velocity of half of the first. The vertical gap between first and second ball is 15m at t=2s.the gap is remained to found constant.the velocities with which the balls were thrown are (g=10) a. 20m/s, 10m/s b. 10,5 c. 16,8 d. 30,15? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A man is standing on top of a building 100m high. He throws 2 balls vertically, one at t=0 and other after a time interval (less than 2).The latte ball is thrown with a velocity of half of the first. The vertical gap between first and second ball is 15m at t=2s.the gap is remained to found constant.the velocities with which the balls were thrown are (g=10) a. 20m/s, 10m/s b. 10,5 c. 16,8 d. 30,15? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A man is standing on top of a building 100m high. He throws 2 balls vertically, one at t=0 and other after a time interval (less than 2).The latte ball is thrown with a velocity of half of the first. The vertical gap between first and second ball is 15m at t=2s.the gap is remained to found constant.the velocities with which the balls were thrown are (g=10) a. 20m/s, 10m/s b. 10,5 c. 16,8 d. 30,15?.
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