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a stone is let to fall from a balloon ascending with an acceleration f. after time t a second stone is dropped. prove that the distance between the stone after time t' since the second stone is dropped is 1/2 (f+g)(t+2t')plz. ans fast....
Most Upvoted Answer
a stone is let to fall from a balloon ascending with an acceleration f...
I think you should first check out the dimensions of the formula you want to prove .it has dimensions of velocity not distance... check it and correct it first
Community Answer
a stone is let to fall from a balloon ascending with an acceleration f...
Proof:
Distance covered by the first stone:
- The distance covered by the first stone dropped from the balloon in time t is given by the formula: d1 = 0.5ft^2
Distance covered by the second stone:
- The distance covered by the second stone dropped after time t' is given by the formula: d2 = 0.5g(t'+t)^2
Total distance between the two stones:
- The total distance between the two stones after time t' is the sum of the distances covered by each stone, which is: d = d1 + d2
- Substituting the values of d1 and d2 into the equation we get: d = 0.5ft^2 + 0.5g(t'+t)^2
Expand the equation:
- Expanding the equation we get: d = 0.5ft^2 + 0.5g(t'^2 + 2tt' + t^2)
- Simplify the equation further: d = 0.5ft^2 + 0.5gt'^2 + gt't + 0.5gt^2
Factor out common terms:
- Factor out the common terms of t and t' to get: d = 0.5(t+2t')(f+g)
Therefore, the distance between the two stones after time t' since the second stone is dropped is 1/2(f+g)(t+2t').
Distance covered by the first stone:
- The distance covered by the first stone dropped from the balloon in time t is given by the formula: d1 = 0.5ft^2
Distance covered by the second stone:
- The distance covered by the second stone dropped after time t' is given by the formula: d2 = 0.5g(t'+t)^2
Total distance between the two stones:
- The total distance between the two stones after time t' is the sum of the distances covered by each stone, which is: d = d1 + d2
- Substituting the values of d1 and d2 into the equation we get: d = 0.5ft^2 + 0.5g(t'+t)^2
Expand the equation:
- Expanding the equation we get: d = 0.5ft^2 + 0.5g(t'^2 + 2tt' + t^2)
- Simplify the equation further: d = 0.5ft^2 + 0.5gt'^2 + gt't + 0.5gt^2
Factor out common terms:
- Factor out the common terms of t and t' to get: d = 0.5(t+2t')(f+g)
Therefore, the distance between the two stones after time t' since the second stone is dropped is 1/2(f+g)(t+2t').
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a stone is let to fall from a balloon ascending with an acceleration f. after time t a second stone is dropped. prove that the distance between the stone after time t' since the second stone is dropped is 1/2 (f+g)(t+2t')plz. ans fast....
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a stone is let to fall from a balloon ascending with an acceleration f. after time t a second stone is dropped. prove that the distance between the stone after time t' since the second stone is dropped is 1/2 (f+g)(t+2t')plz. ans fast.... for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about a stone is let to fall from a balloon ascending with an acceleration f. after time t a second stone is dropped. prove that the distance between the stone after time t' since the second stone is dropped is 1/2 (f+g)(t+2t')plz. ans fast.... covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for a stone is let to fall from a balloon ascending with an acceleration f. after time t a second stone is dropped. prove that the distance between the stone after time t' since the second stone is dropped is 1/2 (f+g)(t+2t')plz. ans fast.....
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