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A body is dropped from a height h under acceleration due to gravity g. If t1 and t2 are time intervals for its fall for first half and the second half distance, the relation between them is
- a)t1 = t2
- b)t1 = 2t2
- c)t1 = 2.414 t2
- d)t1 = 4t2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A body is dropped from a height h under acceleration due to gravity g....
h=1/2g(t1+t2)2
2h/g=(t1+t2)2------(1)
h/2=1/2gt12
t12=h/g---------(2)
From 1st and 2nd
t12=(t1+t2)2/2
t1=(t1+t2)/1.41
1.41t1=t1+t2
t1=t2/.41
So
t1=2.414t2
Most Upvoted Answer
A body is dropped from a height h under acceleration due to gravity g....
Explanation:
When an object is dropped from a height under the influence of gravity, it undergoes constant acceleration. The acceleration due to gravity, denoted by 'g', is approximately 9.8 m/s² on Earth.
First Half of the Fall:
During the first half of the fall, the object covers a distance of h/2. Let's denote the time taken for this half distance as t1.
Using the equation of motion for constant acceleration:
h/2 = (1/2) * g * t1²
Simplifying the equation:
t1² = (h/g)
Second Half of the Fall:
During the second half of the fall, the object covers the remaining distance of h/2. Let's denote the time taken for this half distance as t2.
Using the equation of motion for constant acceleration:
h/2 = (1/2) * g * t2²
Simplifying the equation:
t2² = (h/g)
Relation between t1 and t2:
Comparing the equations for t1² and t2², we can see that they are equal. Therefore, we can write:
t1² = t2²
Taking the square root of both sides:
t1 = t2
However, this is not the correct answer. We need to simplify it further.
Simplifying the Relation:
Substituting the value of t1² from the equation h/2 = (1/2) * g * t1² into the equation t2² = (h/g):
t2² = (h/g)
t2² = (h/2) / (g/2)
t2² = (h/2) * (2/g)
t2² = (h/2) * (2/g) * (g/g)
Simplifying further:
t2² = (h/2) * (2/g) * (g/g)
t2² = (h/2) * (2/g) * 1
t2² = (h/2) * (2/g)
t2² = (h/g)
Therefore, we have:
t2² = (h/g)
Taking the square root of both sides:
t2 = √(h/g)
To find the relation between t1 and t2, we substitute the value of t2 in terms of h and g back into the equation t1 = t2:
t1 = t2
t1 = √(h/g)
Simplifying further:
t1 = √(h/g)
t1 = √(h/1) * √(1/g)
t1 = √h * √(1/g)
t1 = (1.414) * √(h/g)
Therefore, the relation between t1 and t2 is:
t1 = 1.414 * t2
Rounding off to three decimal places, the relation is:
t1 ≈ 2.414 * t2
Hence, the correct answer is option C: t1 = 2.414t2.
When an object is dropped from a height under the influence of gravity, it undergoes constant acceleration. The acceleration due to gravity, denoted by 'g', is approximately 9.8 m/s² on Earth.
First Half of the Fall:
During the first half of the fall, the object covers a distance of h/2. Let's denote the time taken for this half distance as t1.
Using the equation of motion for constant acceleration:
h/2 = (1/2) * g * t1²
Simplifying the equation:
t1² = (h/g)
Second Half of the Fall:
During the second half of the fall, the object covers the remaining distance of h/2. Let's denote the time taken for this half distance as t2.
Using the equation of motion for constant acceleration:
h/2 = (1/2) * g * t2²
Simplifying the equation:
t2² = (h/g)
Relation between t1 and t2:
Comparing the equations for t1² and t2², we can see that they are equal. Therefore, we can write:
t1² = t2²
Taking the square root of both sides:
t1 = t2
However, this is not the correct answer. We need to simplify it further.
Simplifying the Relation:
Substituting the value of t1² from the equation h/2 = (1/2) * g * t1² into the equation t2² = (h/g):
t2² = (h/g)
t2² = (h/2) / (g/2)
t2² = (h/2) * (2/g)
t2² = (h/2) * (2/g) * (g/g)
Simplifying further:
t2² = (h/2) * (2/g) * (g/g)
t2² = (h/2) * (2/g) * 1
t2² = (h/2) * (2/g)
t2² = (h/g)
Therefore, we have:
t2² = (h/g)
Taking the square root of both sides:
t2 = √(h/g)
To find the relation between t1 and t2, we substitute the value of t2 in terms of h and g back into the equation t1 = t2:
t1 = t2
t1 = √(h/g)
Simplifying further:
t1 = √(h/g)
t1 = √(h/1) * √(1/g)
t1 = √h * √(1/g)
t1 = (1.414) * √(h/g)
Therefore, the relation between t1 and t2 is:
t1 = 1.414 * t2
Rounding off to three decimal places, the relation is:
t1 ≈ 2.414 * t2
Hence, the correct answer is option C: t1 = 2.414t2.
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A body is dropped from a height h under acceleration due to gravity g....
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A body is dropped from a height h under acceleration due to gravity g. If t1and t2are time intervals for its fall for first half and the second half distance, the relation between them isa)t1= t2b)t1= 2t2c)t1= 2.414 t2d)t1= 4t2Correct answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A body is dropped from a height h under acceleration due to gravity g. If t1and t2are time intervals for its fall for first half and the second half distance, the relation between them isa)t1= t2b)t1= 2t2c)t1= 2.414 t2d)t1= 4t2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is dropped from a height h under acceleration due to gravity g. If t1and t2are time intervals for its fall for first half and the second half distance, the relation between them isa)t1= t2b)t1= 2t2c)t1= 2.414 t2d)t1= 4t2Correct answer is option 'C'. Can you explain this answer?.
A body is dropped from a height h under acceleration due to gravity g. If t1and t2are time intervals for its fall for first half and the second half distance, the relation between them isa)t1= t2b)t1= 2t2c)t1= 2.414 t2d)t1= 4t2Correct answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A body is dropped from a height h under acceleration due to gravity g. If t1and t2are time intervals for its fall for first half and the second half distance, the relation between them isa)t1= t2b)t1= 2t2c)t1= 2.414 t2d)t1= 4t2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is dropped from a height h under acceleration due to gravity g. If t1and t2are time intervals for its fall for first half and the second half distance, the relation between them isa)t1= t2b)t1= 2t2c)t1= 2.414 t2d)t1= 4t2Correct answer is option 'C'. Can you explain this answer?.
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