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From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. a second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. a third ball released, from the rest from the same location, will reach the ground in s. [jee (main) 2022, 24 june (?
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From the top of a tower, a ball is thrown vertically upward which reac...
Analysis:
Given data:
- Time taken for the first ball to reach the ground when thrown upwards = 6 s
- Time taken for the second ball to reach the ground when thrown downwards = 1.5 s
To solve this problem, we need to consider the motion of the balls in each case and apply the equations of motion.
Ball thrown upwards:
- Let the initial velocity of the ball be u and the acceleration due to gravity be g.
- When the ball reaches the ground, its final velocity is 0 m/s.
- Using the equation of motion: v = u + gt, where v is final velocity, u is initial velocity, g is acceleration due to gravity, and t is time taken.
- When the ball reaches the ground, v = 0 m/s, u = u, and t = 6 s.
- Therefore, 0 = u - 9.8 * 6
- Solving the equation, we get u = 58.8 m/s.
Ball thrown downwards:
- Similar to the first case, let the initial velocity of the ball be u and the acceleration due to gravity be g.
- When the ball reaches the ground, its final velocity is 0 m/s.
- Using the equation of motion: v = u + gt, where v is final velocity, u is initial velocity, g is acceleration due to gravity, and t is time taken.
- When the ball reaches the ground, v = 0 m/s, u = -u (negative because it is thrown downwards), and t = 1.5 s.
- Therefore, 0 = -u - 9.8 * 1.5
- Solving the equation, we get u = -14.7 m/s.
Ball released from rest:
- When the ball is released from rest, its initial velocity u = 0 m/s.
- Using the equation of motion: s = ut + 0.5gt^2, where s is the distance traveled, u is initial velocity, g is acceleration due to gravity, and t is time taken.
- We need to find the time taken for the ball to reach the ground.
- Let's assume the distance traveled by the ball is s meters.
- When the ball reaches the ground, s = 0 (as it reaches the ground).
- Therefore, 0 = 0 + 0.5 * 9.8 * t^2
- Solving the equation, we get t = sqrt(0) = 0 s.
Hence, the third ball released from rest from the same location will reach the ground instantaneously.
Given data:
- Time taken for the first ball to reach the ground when thrown upwards = 6 s
- Time taken for the second ball to reach the ground when thrown downwards = 1.5 s
To solve this problem, we need to consider the motion of the balls in each case and apply the equations of motion.
Ball thrown upwards:
- Let the initial velocity of the ball be u and the acceleration due to gravity be g.
- When the ball reaches the ground, its final velocity is 0 m/s.
- Using the equation of motion: v = u + gt, where v is final velocity, u is initial velocity, g is acceleration due to gravity, and t is time taken.
- When the ball reaches the ground, v = 0 m/s, u = u, and t = 6 s.
- Therefore, 0 = u - 9.8 * 6
- Solving the equation, we get u = 58.8 m/s.
Ball thrown downwards:
- Similar to the first case, let the initial velocity of the ball be u and the acceleration due to gravity be g.
- When the ball reaches the ground, its final velocity is 0 m/s.
- Using the equation of motion: v = u + gt, where v is final velocity, u is initial velocity, g is acceleration due to gravity, and t is time taken.
- When the ball reaches the ground, v = 0 m/s, u = -u (negative because it is thrown downwards), and t = 1.5 s.
- Therefore, 0 = -u - 9.8 * 1.5
- Solving the equation, we get u = -14.7 m/s.
Ball released from rest:
- When the ball is released from rest, its initial velocity u = 0 m/s.
- Using the equation of motion: s = ut + 0.5gt^2, where s is the distance traveled, u is initial velocity, g is acceleration due to gravity, and t is time taken.
- We need to find the time taken for the ball to reach the ground.
- Let's assume the distance traveled by the ball is s meters.
- When the ball reaches the ground, s = 0 (as it reaches the ground).
- Therefore, 0 = 0 + 0.5 * 9.8 * t^2
- Solving the equation, we get t = sqrt(0) = 0 s.
Hence, the third ball released from rest from the same location will reach the ground instantaneously.
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From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. a second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. a third ball released, from the rest from the same location, will reach the ground in s. [jee (main) 2022, 24 june (?
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From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. a second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. a third ball released, from the rest from the same location, will reach the ground in s. [jee (main) 2022, 24 june (? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. a second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. a third ball released, from the rest from the same location, will reach the ground in s. [jee (main) 2022, 24 june (? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. a second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. a third ball released, from the rest from the same location, will reach the ground in s. [jee (main) 2022, 24 june (?.
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