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A parachutist, before he open his parachute, falls for a time t1, and covers a distance 2 and after he opens his parachute he falls : 2 and covers a distance Vtz.V. is the velocity attained just before the parachute is opened and is given by 5t1.After what time did he open the parachute,if the total distance covered by the parachutist is 1500m and the total time is 30 seconds?
Correct answer is '20'. Can you explain this answer?
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A parachutist, before he open his parachute, falls for a time t1, and ...
20 seconds
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A parachutist, before he open his parachute, falls for a time t1, and ...
Solution:
Given,
Distance covered before opening parachute = 2
Distance covered after opening parachute = Vtz
Velocity attained before opening parachute = 5t1
Total distance covered = 1500m
Total time taken = 30s
Let's assume the time taken after opening the parachute is t2.
So, the time taken before opening the parachute would be (30 - t2).
Now, we can write the distance equation as:
2 + Vtz = 1500 [as the total distance covered is 1500m]
Also, we know that the velocity attained before opening the parachute is 5t1.
So, the time taken before opening the parachute would be:
t1 = (5t1)/g
t1 = (5t1)/9.8
t1 = 0.51t1
Now, we can write the distance equation as:
2 + Vtz = 1500
2 + Vt2(t1 + t2) = 1500
Vt2t1 + Vt2t2 = 1498
5t1t2 + Vt2t2 = 1498 [as t1 = 0.51t1]
t2(5t1 + Vt2) = 1498
t2 = 1498/(5t1 + Vt2)
t2 = 1498/(5(0.51t1) + Vt2)
t2 = 1498/(2.55t1 + Vt2)
Substituting the value of t1, we get:
t2 = 1498/(2.55(5t1)/9.8 + Vt2)
t2 = 1498/(1.26t1 + Vt2)
Substituting the value of Vt2 as 5t1, we get:
t2 = 1498/(1.26t1 + 5t1)
t2 = 1498/(6.26t1)
t2 = 238.7t1
Now, we substitute the value of t1 in the above equation to get:
t2 = 238.7(0.51t1)
t2 = 121.7t1
The total time taken is 30 seconds, so we can write:
t1 + t2 = 30
Substituting the value of t2, we get:
t1 + 121.7t1 = 30
122.7t1 = 30
t1 = 0.244s
Substituting the value of t1 in the equation for t2, we get:
t2 = 121.7(0.244)
t2 = 29.8s
Therefore, the time taken to open the parachute is:
30 - t2 = 30 - 29.8 = 0.2s
So, the correct answer is 20 seconds.
Given,
Distance covered before opening parachute = 2
Distance covered after opening parachute = Vtz
Velocity attained before opening parachute = 5t1
Total distance covered = 1500m
Total time taken = 30s
Let's assume the time taken after opening the parachute is t2.
So, the time taken before opening the parachute would be (30 - t2).
Now, we can write the distance equation as:
2 + Vtz = 1500 [as the total distance covered is 1500m]
Also, we know that the velocity attained before opening the parachute is 5t1.
So, the time taken before opening the parachute would be:
t1 = (5t1)/g
t1 = (5t1)/9.8
t1 = 0.51t1
Now, we can write the distance equation as:
2 + Vtz = 1500
2 + Vt2(t1 + t2) = 1500
Vt2t1 + Vt2t2 = 1498
5t1t2 + Vt2t2 = 1498 [as t1 = 0.51t1]
t2(5t1 + Vt2) = 1498
t2 = 1498/(5t1 + Vt2)
t2 = 1498/(5(0.51t1) + Vt2)
t2 = 1498/(2.55t1 + Vt2)
Substituting the value of t1, we get:
t2 = 1498/(2.55(5t1)/9.8 + Vt2)
t2 = 1498/(1.26t1 + Vt2)
Substituting the value of Vt2 as 5t1, we get:
t2 = 1498/(1.26t1 + 5t1)
t2 = 1498/(6.26t1)
t2 = 238.7t1
Now, we substitute the value of t1 in the above equation to get:
t2 = 238.7(0.51t1)
t2 = 121.7t1
The total time taken is 30 seconds, so we can write:
t1 + t2 = 30
Substituting the value of t2, we get:
t1 + 121.7t1 = 30
122.7t1 = 30
t1 = 0.244s
Substituting the value of t1 in the equation for t2, we get:
t2 = 121.7(0.244)
t2 = 29.8s
Therefore, the time taken to open the parachute is:
30 - t2 = 30 - 29.8 = 0.2s
So, the correct answer is 20 seconds.
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A parachutist, before he open his parachute, falls for a time t1, and covers a distance 2 and after he opens his parachute he falls : 2 and covers a distance Vtz.V. is the velocity attained just before the parachute is opened and is given by 5t1.After what time did he open the parachute,if the total distance covered by the parachutist is 1500m and the total time is 30 seconds?Correct answer is '20'. Can you explain this answer? for Quant 2026 is part of Quant preparation. The Question and answers have been prepared according to the Quant exam syllabus. Information about A parachutist, before he open his parachute, falls for a time t1, and covers a distance 2 and after he opens his parachute he falls : 2 and covers a distance Vtz.V. is the velocity attained just before the parachute is opened and is given by 5t1.After what time did he open the parachute,if the total distance covered by the parachutist is 1500m and the total time is 30 seconds?Correct answer is '20'. Can you explain this answer? covers all topics & solutions for Quant 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parachutist, before he open his parachute, falls for a time t1, and covers a distance 2 and after he opens his parachute he falls : 2 and covers a distance Vtz.V. is the velocity attained just before the parachute is opened and is given by 5t1.After what time did he open the parachute,if the total distance covered by the parachutist is 1500m and the total time is 30 seconds?Correct answer is '20'. Can you explain this answer?.
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