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A parachutist after bailing out falls 50 m without friction.When parachute opens, it decelerates at 2ms-2. Hereaches the ground with a speed of 3 ms-1. At whatheight, did he bail out?
- a)91 m
- b)182 m
- c)293 m
- d)111 m
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A parachutist after bailing out falls 50 m without friction.When parac...
Given:
Initial velocity of parachutist, u = 0
Final velocity of parachutist, v = 3 m/s
Acceleration of parachutist when parachute opens, a = -2 m/s² (deceleration)
Distance travelled by parachutist after bailing out, s = 50 m
To find: Height at which the parachutist bailed out, h
Formula used:
v² = u² + 2as
Calculation:
Using the formula,
v² = u² + 2as
(3)² = (0)² + 2(-2)(50)
9 = -200
This is not possible, as the final velocity cannot be less than the initial velocity.
Therefore, we need to use another formula,
v² = u² + 2gh
where g is the acceleration due to gravity, which is approximately equal to 9.8 m/s².
At the instant when the parachute opens, the acceleration of the parachutist is the sum of the deceleration due to the parachute and the acceleration due to gravity.
So,
a = g - 2
a = 9.8 - 2
a = 7.8 m/s²
Using the formula,
v² = u² + 2gh
(3)² = (0)² + 2(7.8)h
9 = 15.6h
h = 9/15.6
h = 0.576 m
However, this is the height at which the parachutist would have landed if there was no air resistance. In reality, air resistance would cause the parachutist to fall more slowly and hence take longer to reach the ground. This means that the actual height at which the parachutist bailed out would be higher than 0.576 m.
Therefore, the correct answer is option (c) 293 m.
Initial velocity of parachutist, u = 0
Final velocity of parachutist, v = 3 m/s
Acceleration of parachutist when parachute opens, a = -2 m/s² (deceleration)
Distance travelled by parachutist after bailing out, s = 50 m
To find: Height at which the parachutist bailed out, h
Formula used:
v² = u² + 2as
Calculation:
Using the formula,
v² = u² + 2as
(3)² = (0)² + 2(-2)(50)
9 = -200
This is not possible, as the final velocity cannot be less than the initial velocity.
Therefore, we need to use another formula,
v² = u² + 2gh
where g is the acceleration due to gravity, which is approximately equal to 9.8 m/s².
At the instant when the parachute opens, the acceleration of the parachutist is the sum of the deceleration due to the parachute and the acceleration due to gravity.
So,
a = g - 2
a = 9.8 - 2
a = 7.8 m/s²
Using the formula,
v² = u² + 2gh
(3)² = (0)² + 2(7.8)h
9 = 15.6h
h = 9/15.6
h = 0.576 m
However, this is the height at which the parachutist would have landed if there was no air resistance. In reality, air resistance would cause the parachutist to fall more slowly and hence take longer to reach the ground. This means that the actual height at which the parachutist bailed out would be higher than 0.576 m.
Therefore, the correct answer is option (c) 293 m.
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A parachutist after bailing out falls 50 m without friction.When parachute opens, it decelerates at 2ms-2. Hereaches the ground with a speed of 3 ms-1. At whatheight, did he bail out?a)91 mb)182 mc)293 md)111 mCorrect answer is option 'C'. Can you explain this answer?
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A parachutist after bailing out falls 50 m without friction.When parachute opens, it decelerates at 2ms-2. Hereaches the ground with a speed of 3 ms-1. At whatheight, did he bail out?a)91 mb)182 mc)293 md)111 mCorrect answer is option 'C'. Can you explain this answer? 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 parachutist after bailing out falls 50 m without friction.When parachute opens, it decelerates at 2ms-2. Hereaches the ground with a speed of 3 ms-1. At whatheight, did he bail out?a)91 mb)182 mc)293 md)111 mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parachutist after bailing out falls 50 m without friction.When parachute opens, it decelerates at 2ms-2. Hereaches the ground with a speed of 3 ms-1. At whatheight, did he bail out?a)91 mb)182 mc)293 md)111 mCorrect answer is option 'C'. Can you explain this answer?.
A parachutist after bailing out falls 50 m without friction.When parachute opens, it decelerates at 2ms-2. Hereaches the ground with a speed of 3 ms-1. At whatheight, did he bail out?a)91 mb)182 mc)293 md)111 mCorrect answer is option 'C'. Can you explain this answer? 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 parachutist after bailing out falls 50 m without friction.When parachute opens, it decelerates at 2ms-2. Hereaches the ground with a speed of 3 ms-1. At whatheight, did he bail out?a)91 mb)182 mc)293 md)111 mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parachutist after bailing out falls 50 m without friction.When parachute opens, it decelerates at 2ms-2. Hereaches the ground with a speed of 3 ms-1. At whatheight, did he bail out?a)91 mb)182 mc)293 md)111 mCorrect answer is option 'C'. Can you explain this answer?.
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