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A ball is thrown vertically upwards with an initial speed of 20m/s . Two seconds later , as stone is thrown vertically( from the same initial height as of the ball) with an initial speed 24m/s. At what height above the release point will the ball and the stone pass each other?
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A ball is thrown vertically upwards with an initial speed of 20m/s . T...
Introduction:
We have a ball and a stone thrown vertically upwards from the same initial height. The ball is thrown with an initial speed of 20m/s, while the stone is thrown 2 seconds later with an initial speed of 24m/s. We need to determine the height above the release point where the ball and the stone pass each other.
Step 1: Analyzing the motion of the ball:
When the ball is thrown vertically upwards, we can analyze its motion using the equations of motion. The initial velocity of the ball is 20m/s, and the acceleration due to gravity is -9.8m/s² (negative because it acts in the opposite direction to the initial velocity).
Using the equation v = u + at, where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
We can calculate that the ball will reach its maximum height after 2 seconds. The final velocity at this point will be 0m/s.
Step 2: Determining the maximum height reached by the ball:
Using the equation v² = u² + 2as, where:
- s is the distance traveled
- a is the acceleration
- u is the initial velocity
- v is the final velocity (which is 0m/s at the maximum height)
We can rearrange the equation to solve for s:
s = (v² - u²) / (2a)
Substituting the given values, we find that the maximum height reached by the ball is:
s = (0² - 20²) / (2 * -9.8)
s = -400 / -19.6
s ≈ 20.41m
Step 3: Analyzing the motion of the stone:
The stone is thrown 2 seconds after the ball with an initial speed of 24m/s. Since the stone is thrown vertically upwards, it will also follow the same path as the ball.
Step 4: Determining the height where the ball and stone pass each other:
We know that the stone is thrown 2 seconds after the ball. During this 2-second interval, the ball would have traveled upwards and reached its maximum height. Thus, when the stone is thrown, it will be at the same height as the maximum height reached by the ball.
Therefore, the height above the release point where the ball and the stone pass each other is approximately 20.41m, which is the maximum height reached by the ball.
Conclusion:
The ball and the stone pass each other at a height of approximately 20.41m above the release point. The stone is thrown 2 seconds after the ball, and during this time, the ball reaches its maximum height. Both the ball and the stone follow the same path, so they meet at the point where the ball reaches its maximum height.
We have a ball and a stone thrown vertically upwards from the same initial height. The ball is thrown with an initial speed of 20m/s, while the stone is thrown 2 seconds later with an initial speed of 24m/s. We need to determine the height above the release point where the ball and the stone pass each other.
Step 1: Analyzing the motion of the ball:
When the ball is thrown vertically upwards, we can analyze its motion using the equations of motion. The initial velocity of the ball is 20m/s, and the acceleration due to gravity is -9.8m/s² (negative because it acts in the opposite direction to the initial velocity).
Using the equation v = u + at, where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
We can calculate that the ball will reach its maximum height after 2 seconds. The final velocity at this point will be 0m/s.
Step 2: Determining the maximum height reached by the ball:
Using the equation v² = u² + 2as, where:
- s is the distance traveled
- a is the acceleration
- u is the initial velocity
- v is the final velocity (which is 0m/s at the maximum height)
We can rearrange the equation to solve for s:
s = (v² - u²) / (2a)
Substituting the given values, we find that the maximum height reached by the ball is:
s = (0² - 20²) / (2 * -9.8)
s = -400 / -19.6
s ≈ 20.41m
Step 3: Analyzing the motion of the stone:
The stone is thrown 2 seconds after the ball with an initial speed of 24m/s. Since the stone is thrown vertically upwards, it will also follow the same path as the ball.
Step 4: Determining the height where the ball and stone pass each other:
We know that the stone is thrown 2 seconds after the ball. During this 2-second interval, the ball would have traveled upwards and reached its maximum height. Thus, when the stone is thrown, it will be at the same height as the maximum height reached by the ball.
Therefore, the height above the release point where the ball and the stone pass each other is approximately 20.41m, which is the maximum height reached by the ball.
Conclusion:
The ball and the stone pass each other at a height of approximately 20.41m above the release point. The stone is thrown 2 seconds after the ball, and during this time, the ball reaches its maximum height. Both the ball and the stone follow the same path, so they meet at the point where the ball reaches its maximum height.
Community Answer
A ball is thrown vertically upwards with an initial speed of 20m/s . T...
yep I got 16.55 m the time is 2.83 sec actually I did a silly mistake of sign change
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A ball is thrown vertically upwards with an initial speed of 20m/s . Two seconds later , as stone is thrown vertically( from the same initial height as of the ball) with an initial speed 24m/s. At what height above the release point will the ball and the stone pass each other?
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A ball is thrown vertically upwards with an initial speed of 20m/s . Two seconds later , as stone is thrown vertically( from the same initial height as of the ball) with an initial speed 24m/s. At what height above the release point will the ball and the stone pass each other? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A ball is thrown vertically upwards with an initial speed of 20m/s . Two seconds later , as stone is thrown vertically( from the same initial height as of the ball) with an initial speed 24m/s. At what height above the release point will the ball and the stone pass each other? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is thrown vertically upwards with an initial speed of 20m/s . Two seconds later , as stone is thrown vertically( from the same initial height as of the ball) with an initial speed 24m/s. At what height above the release point will the ball and the stone pass each other?.
A ball is thrown vertically upwards with an initial speed of 20m/s . Two seconds later , as stone is thrown vertically( from the same initial height as of the ball) with an initial speed 24m/s. At what height above the release point will the ball and the stone pass each other? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A ball is thrown vertically upwards with an initial speed of 20m/s . Two seconds later , as stone is thrown vertically( from the same initial height as of the ball) with an initial speed 24m/s. At what height above the release point will the ball and the stone pass each other? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball is thrown vertically upwards with an initial speed of 20m/s . Two seconds later , as stone is thrown vertically( from the same initial height as of the ball) with an initial speed 24m/s. At what height above the release point will the ball and the stone pass each other?.
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