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A parachutist after bailing out falls 50 m without friction.
Wh en par achute open s, it decelerates at 2 m /s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?
Wh en par achute open s, it decelerates at 2 m /s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?
- a)182 m
- b)91 m
- c)111 m
- d)293 m
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A parachutist after bailing out falls 50 m without friction.Wh en par ...
The velocity of parachutist when parachute opens is
Initially he has fallen 50 m.
∴ Total height from where he bailed out = 243 + 50 = 293 m
∴ Total height from where he bailed out = 243 + 50 = 293 m
Most Upvoted Answer
A parachutist after bailing out falls 50 m without friction.Wh en par ...
Given data:
- Initial velocity of the parachutist = 0 m/s
- Acceleration due to gravity = 9.8 m/s^2
- Deceleration when parachute opens = 2 m/s^2
- Final velocity before reaching the ground = 3 m/s
Calculating the height where the parachutist bailed out:
- Using the first equation of motion, we can calculate the distance fallen without friction:
h = (1/2) * g * t^2
h = (1/2) * 9.8 * 2^2
h = 19.6 m
- Using the third equation of motion to calculate the time taken for the parachutist to reach the ground after the parachute opens:
3^2 = 0^2 + 2 * (-9.8) * d
d = 0.459 m
3 = 0 + 2 * (-2) * t
t = 0.75 s
- Calculating the total time taken for the parachutist to reach the ground:
Total time = Time taken to fall without friction + Time taken after parachute opens
Total time = 2 + 0.75
Total time = 2.75 s
- Using the first equation of motion to calculate the initial height where the parachutist bailed out:
h = 0 * 2.75 + (1/2) * (-2) * 2.75^2
h = 293 m
Therefore, the height at which the parachutist bailed out was 293 m. Hence, the correct answer is option 'D'.
- Initial velocity of the parachutist = 0 m/s
- Acceleration due to gravity = 9.8 m/s^2
- Deceleration when parachute opens = 2 m/s^2
- Final velocity before reaching the ground = 3 m/s
Calculating the height where the parachutist bailed out:
- Using the first equation of motion, we can calculate the distance fallen without friction:
h = (1/2) * g * t^2
h = (1/2) * 9.8 * 2^2
h = 19.6 m
- Using the third equation of motion to calculate the time taken for the parachutist to reach the ground after the parachute opens:
3^2 = 0^2 + 2 * (-9.8) * d
d = 0.459 m
3 = 0 + 2 * (-2) * t
t = 0.75 s
- Calculating the total time taken for the parachutist to reach the ground:
Total time = Time taken to fall without friction + Time taken after parachute opens
Total time = 2 + 0.75
Total time = 2.75 s
- Using the first equation of motion to calculate the initial height where the parachutist bailed out:
h = 0 * 2.75 + (1/2) * (-2) * 2.75^2
h = 293 m
Therefore, the height at which the parachutist bailed out was 293 m. Hence, the correct answer is option 'D'.
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A parachutist after bailing out falls 50 m without friction.Wh en par achute open s, it decelerates at 2 m /s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?a)182 mb)91 mc)111 md)293 mCorrect answer is option 'D'. Can you explain this answer?
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A parachutist after bailing out falls 50 m without friction.Wh en par achute open s, it decelerates at 2 m /s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?a)182 mb)91 mc)111 md)293 mCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A parachutist after bailing out falls 50 m without friction.Wh en par achute open s, it decelerates at 2 m /s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?a)182 mb)91 mc)111 md)293 mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parachutist after bailing out falls 50 m without friction.Wh en par achute open s, it decelerates at 2 m /s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?a)182 mb)91 mc)111 md)293 mCorrect answer is option 'D'. Can you explain this answer?.
A parachutist after bailing out falls 50 m without friction.Wh en par achute open s, it decelerates at 2 m /s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?a)182 mb)91 mc)111 md)293 mCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A parachutist after bailing out falls 50 m without friction.Wh en par achute open s, it decelerates at 2 m /s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?a)182 mb)91 mc)111 md)293 mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parachutist after bailing out falls 50 m without friction.Wh en par achute open s, it decelerates at 2 m /s2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?a)182 mb)91 mc)111 md)293 mCorrect answer is option 'D'. Can you explain this answer?.
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