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A body is projected vertically upwards with a velocity u and time t seconds later .another body is projected vertically upwards from the same point with same velocity from the ground.they meet at a height of ? 1) u^2/16 g. 2)u^2/32g. 3)u^2/2g gt^2/8. 4)u^2/2g - gt^2/8. The answer for this is u^2/2g - gt^2/8.please explain this.?
Most Upvoted Answer
A body is projected vertically upwards with a velocity u and time t se...
Solution:
1. Calculation of height reached by the first body:
Using the equation of motion, we can find the height reached by the first body after time t seconds as:
s1 = ut - 1/2 gt^2
2. Calculation of time taken by the second body to reach the same height:
Let the time taken by the second body to reach the same height be t1 seconds. Using the equation of motion, we can find the height reached by the second body after time t1 seconds as:
s2 = u t1 - 1/2 g t1^2
Since both bodies reach the same height at the same time, we can equate s1 and s2 to get:
ut - 1/2 gt^2 = u t1 - 1/2 g t1^2
Simplifying the above equation, we get:
t1 = t - (u/g)
3. Calculation of height reached by the second body:
Using the value of t1, we can find the height reached by the second body as:
s2 = u(t - u/g) - 1/2 g (t - u/g)^2
Simplifying the above equation, we get:
s2 = ut - 1/2 gt^2 + u^2/2g - gt^2/8
4. Height at which both bodies meet:
Since both bodies meet at the same height, we can equate s1 and s2 to get:
ut - 1/2 gt^2 = ut - 1/2 gt^2 + u^2/2g - gt^2/8
Simplifying the above equation, we get:
Height at which both bodies meet = u^2/2g - gt^2/8
Thus, the correct option is 4) u^2/2g - gt^2/8.
1. Calculation of height reached by the first body:
Using the equation of motion, we can find the height reached by the first body after time t seconds as:
s1 = ut - 1/2 gt^2
2. Calculation of time taken by the second body to reach the same height:
Let the time taken by the second body to reach the same height be t1 seconds. Using the equation of motion, we can find the height reached by the second body after time t1 seconds as:
s2 = u t1 - 1/2 g t1^2
Since both bodies reach the same height at the same time, we can equate s1 and s2 to get:
ut - 1/2 gt^2 = u t1 - 1/2 g t1^2
Simplifying the above equation, we get:
t1 = t - (u/g)
3. Calculation of height reached by the second body:
Using the value of t1, we can find the height reached by the second body as:
s2 = u(t - u/g) - 1/2 g (t - u/g)^2
Simplifying the above equation, we get:
s2 = ut - 1/2 gt^2 + u^2/2g - gt^2/8
4. Height at which both bodies meet:
Since both bodies meet at the same height, we can equate s1 and s2 to get:
ut - 1/2 gt^2 = ut - 1/2 gt^2 + u^2/2g - gt^2/8
Simplifying the above equation, we get:
Height at which both bodies meet = u^2/2g - gt^2/8
Thus, the correct option is 4) u^2/2g - gt^2/8.
Community Answer
A body is projected vertically upwards with a velocity u and time t se...
The answer is from the ground reference.. Ok? So see it as from the ground...
First, take the bodies as A and B for easiness... 'A' moves up first...
As it's given in the question that the bodies meet each other despite the fact that B is thrown later, clearly means that the body A goes its maximum height and then meets B while returning... Ok?
So, calculate the DISTANCE (not the displacement) S= (u^2/2g) + (ut+1/2gt^2) = u^2/2g + 1/2gt^2 {since u from top is 0)The second part which is ut+1/2...... Is the distance covered while returning so it's clearly the displacement measured from the top point... Okay?
So, you can calculate that the maximum height is u^2/2g and the displacement (of A while returning, measured from the top) is 1/2gt^2... So, the distance from the ground to the meeting point is u^2/2g - 1/2gt^2
I hope this gives you an explanation you were finding... If you have ANY doubt on this, you can surely ask... :)
First, take the bodies as A and B for easiness... 'A' moves up first...
As it's given in the question that the bodies meet each other despite the fact that B is thrown later, clearly means that the body A goes its maximum height and then meets B while returning... Ok?
So, calculate the DISTANCE (not the displacement) S= (u^2/2g) + (ut+1/2gt^2) = u^2/2g + 1/2gt^2 {since u from top is 0)The second part which is ut+1/2...... Is the distance covered while returning so it's clearly the displacement measured from the top point... Okay?
So, you can calculate that the maximum height is u^2/2g and the displacement (of A while returning, measured from the top) is 1/2gt^2... So, the distance from the ground to the meeting point is u^2/2g - 1/2gt^2
I hope this gives you an explanation you were finding... If you have ANY doubt on this, you can surely ask... :)
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A body is projected vertically upwards with a velocity u and time t seconds later .another body is projected vertically upwards from the same point with same velocity from the ground.they meet at a height of ? 1) u^2/16 g. 2)u^2/32g. 3)u^2/2g gt^2/8. 4)u^2/2g - gt^2/8. The answer for this is u^2/2g - gt^2/8.please explain this.?
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A body is projected vertically upwards with a velocity u and time t seconds later .another body is projected vertically upwards from the same point with same velocity from the ground.they meet at a height of ? 1) u^2/16 g. 2)u^2/32g. 3)u^2/2g gt^2/8. 4)u^2/2g - gt^2/8. The answer for this is u^2/2g - gt^2/8.please explain this.? 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 body is projected vertically upwards with a velocity u and time t seconds later .another body is projected vertically upwards from the same point with same velocity from the ground.they meet at a height of ? 1) u^2/16 g. 2)u^2/32g. 3)u^2/2g gt^2/8. 4)u^2/2g - gt^2/8. The answer for this is u^2/2g - gt^2/8.please explain this.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is projected vertically upwards with a velocity u and time t seconds later .another body is projected vertically upwards from the same point with same velocity from the ground.they meet at a height of ? 1) u^2/16 g. 2)u^2/32g. 3)u^2/2g gt^2/8. 4)u^2/2g - gt^2/8. The answer for this is u^2/2g - gt^2/8.please explain this.?.
A body is projected vertically upwards with a velocity u and time t seconds later .another body is projected vertically upwards from the same point with same velocity from the ground.they meet at a height of ? 1) u^2/16 g. 2)u^2/32g. 3)u^2/2g gt^2/8. 4)u^2/2g - gt^2/8. The answer for this is u^2/2g - gt^2/8.please explain this.? 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 body is projected vertically upwards with a velocity u and time t seconds later .another body is projected vertically upwards from the same point with same velocity from the ground.they meet at a height of ? 1) u^2/16 g. 2)u^2/32g. 3)u^2/2g gt^2/8. 4)u^2/2g - gt^2/8. The answer for this is u^2/2g - gt^2/8.please explain this.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body is projected vertically upwards with a velocity u and time t seconds later .another body is projected vertically upwards from the same point with same velocity from the ground.they meet at a height of ? 1) u^2/16 g. 2)u^2/32g. 3)u^2/2g gt^2/8. 4)u^2/2g - gt^2/8. The answer for this is u^2/2g - gt^2/8.please explain this.?.
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