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A small ball is thrown between 2 vertical walls such that in the absence of the wall.it's range would have been 5d .the angle of projection is a.given that all the collisions are perfectly elastic.find total no of collisions before the balls comes to back ground.?
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A small ball is thrown between 2 vertical walls such that in the absen...
Total number of collision with the walls before the ball comes back to the ground are nine.
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A small ball is thrown between 2 vertical walls such that in the absen...
Analysis of the Problem:
- The ball is thrown between two vertical walls.
- In the absence of the walls, the range of the ball would have been 5d, where d is the distance between the walls.
- The angle of projection is given as 'a'.
- All collisions are perfectly elastic.
- We need to find the total number of collisions before the ball comes to the ground.

Solution:
To solve this problem, we need to understand the motion of the ball and how it bounces off the walls.

Motion of the Ball:
1. When the ball is thrown, it follows a parabolic trajectory due to the angle of projection.
2. It reaches a maximum height and then starts coming down towards the ground.
3. The ball bounces off the walls elastically, which means there is no loss of kinetic energy during collisions.
4. The ball will continue to bounce between the walls until it reaches the ground.

Range of the Ball:
The range of the ball in the absence of the walls is given as 5d. Range is the horizontal distance covered by the ball.
- Range = (initial velocity)^2 * sin(2a) / g
- In the absence of the walls, the initial velocity of the ball remains the same. Therefore, the range is directly proportional to sin(2a).
- If the walls were not there, the range would have been 5d. So, sin(2a) = 5d/Range.

Number of Collisions:
To find the number of collisions, we need to consider the motion of the ball and the time it takes to cover the distance between the walls.

1. The time taken by the ball to cover the distance between the walls is given by:
- Time taken = (2 * distance) / horizontal velocity
- As the walls are vertical, the horizontal velocity remains constant throughout the motion.
- Therefore, the time taken between each collision is constant.

2. The total time taken by the ball to reach the ground can be calculated using the equation of motion:
- Vertical distance covered = initial velocity * time - (0.5 * g * time^2)
- The total time taken to reach the ground can be found by equating the above equation to zero and solving for time.

3. The total number of collisions can be calculated by dividing the total time taken to reach the ground by the time taken between each collision.

Summary:
By analyzing the motion of the ball, considering the range, and calculating the time taken, we can find the total number of collisions before the ball comes to the ground.
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A small ball is thrown between 2 vertical walls such that in the absence of the wall.it's range would have been 5d .the angle of projection is a.given that all the collisions are perfectly elastic.find total no of collisions before the balls comes to back ground.?
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A small ball is thrown between 2 vertical walls such that in the absence of the wall.it's range would have been 5d .the angle of projection is a.given that all the collisions are perfectly elastic.find total no of collisions before the balls comes to back ground.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A small ball is thrown between 2 vertical walls such that in the absence of the wall.it's range would have been 5d .the angle of projection is a.given that all the collisions are perfectly elastic.find total no of collisions before the balls comes to back ground.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A small ball is thrown between 2 vertical walls such that in the absence of the wall.it's range would have been 5d .the angle of projection is a.given that all the collisions are perfectly elastic.find total no of collisions before the balls comes to back ground.?.
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