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A body at rest is dropped from a height h and strikes the ground with a velocity 3 m/s. Another body of same mass is dropped from the same height with an initial velocity of 4 m/s. The final velocity of second mass, with which it strikes the ground, is (take g=10 m/s2)
- a)3 m/s
- b)4 m/s
- c)5 m/s
- d)8 m/s
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A body at rest is dropped from a height h and strikes the ground with ...
Using v2 = u2 + 2 as
(3)2 = 0 + 2 g x s or 2 gs = 9
when initial velocity is 4 m/s, then
v2 = 42 + 2 gs = 16 + 9 = 25 or v = 5 m/s.
(3)2 = 0 + 2 g x s or 2 gs = 9
when initial velocity is 4 m/s, then
v2 = 42 + 2 gs = 16 + 9 = 25 or v = 5 m/s.
Most Upvoted Answer
A body at rest is dropped from a height h and strikes the ground with ...
Dropping bodies from a height
When a body is dropped from a height, it falls freely under the influence of gravity. The acceleration due to gravity, denoted as 'g', is a constant value of 10 m/s^2 on the surface of the Earth.
First body:
Given that the first body is dropped from a height 'h' and strikes the ground with a velocity of 3 m/s.
Since the body falls freely, it undergoes constant acceleration due to gravity. We can use the equation of motion to determine the final velocity of the first body.
The equation of motion for the first body can be written as:
v^2 = u^2 + 2as
where,
v = final velocity of the first body
u = initial velocity of the first body (0 m/s as it is dropped from rest)
a = acceleration due to gravity (10 m/s^2)
s = distance fallen (h)
Substituting the given values into the equation, we get:
v^2 = 0^2 + 2 * 10 * h
v^2 = 20h
v = sqrt(20h)
Since the final velocity is given as 3 m/s, we have:
3 = sqrt(20h)
9 = 20h
h = 9/20
Second body:
Now, let's consider the second body, which is also dropped from the same height 'h' but with an initial velocity of 4 m/s.
Using the same equation of motion, we can determine the final velocity of the second body.
v^2 = u^2 + 2as
Substituting the given values into the equation, we get:
v^2 = 4^2 + 2 * 10 * h
v^2 = 16 + 20h
v = sqrt(16 + 20h)
Substituting the value of h we found earlier, we have:
v = sqrt(16 + 20 * 9/20)
v = sqrt(16 + 9)
v = sqrt(25)
v = 5 m/s
Conclusion:
Therefore, the final velocity of the second body when it strikes the ground is 5 m/s, which corresponds to option C.
When a body is dropped from a height, it falls freely under the influence of gravity. The acceleration due to gravity, denoted as 'g', is a constant value of 10 m/s^2 on the surface of the Earth.
First body:
Given that the first body is dropped from a height 'h' and strikes the ground with a velocity of 3 m/s.
Since the body falls freely, it undergoes constant acceleration due to gravity. We can use the equation of motion to determine the final velocity of the first body.
The equation of motion for the first body can be written as:
v^2 = u^2 + 2as
where,
v = final velocity of the first body
u = initial velocity of the first body (0 m/s as it is dropped from rest)
a = acceleration due to gravity (10 m/s^2)
s = distance fallen (h)
Substituting the given values into the equation, we get:
v^2 = 0^2 + 2 * 10 * h
v^2 = 20h
v = sqrt(20h)
Since the final velocity is given as 3 m/s, we have:
3 = sqrt(20h)
9 = 20h
h = 9/20
Second body:
Now, let's consider the second body, which is also dropped from the same height 'h' but with an initial velocity of 4 m/s.
Using the same equation of motion, we can determine the final velocity of the second body.
v^2 = u^2 + 2as
Substituting the given values into the equation, we get:
v^2 = 4^2 + 2 * 10 * h
v^2 = 16 + 20h
v = sqrt(16 + 20h)
Substituting the value of h we found earlier, we have:
v = sqrt(16 + 20 * 9/20)
v = sqrt(16 + 9)
v = sqrt(25)
v = 5 m/s
Conclusion:
Therefore, the final velocity of the second body when it strikes the ground is 5 m/s, which corresponds to option C.
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A body at rest is dropped from a height h and strikes the ground with a velocity 3 m/s. Another body of same mass is dropped from the same height with an initial velocity of 4 m/s. The final velocity of second mass, with which it strikes the ground, is (take g=10 m/s2)a)3 m/sb)4 m/sc)5 m/sd)8 m/sCorrect answer is option 'C'. Can you explain this answer?
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A body at rest is dropped from a height h and strikes the ground with a velocity 3 m/s. Another body of same mass is dropped from the same height with an initial velocity of 4 m/s. The final velocity of second mass, with which it strikes the ground, is (take g=10 m/s2)a)3 m/sb)4 m/sc)5 m/sd)8 m/sCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A body at rest is dropped from a height h and strikes the ground with a velocity 3 m/s. Another body of same mass is dropped from the same height with an initial velocity of 4 m/s. The final velocity of second mass, with which it strikes the ground, is (take g=10 m/s2)a)3 m/sb)4 m/sc)5 m/sd)8 m/sCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body at rest is dropped from a height h and strikes the ground with a velocity 3 m/s. Another body of same mass is dropped from the same height with an initial velocity of 4 m/s. The final velocity of second mass, with which it strikes the ground, is (take g=10 m/s2)a)3 m/sb)4 m/sc)5 m/sd)8 m/sCorrect answer is option 'C'. Can you explain this answer?.
A body at rest is dropped from a height h and strikes the ground with a velocity 3 m/s. Another body of same mass is dropped from the same height with an initial velocity of 4 m/s. The final velocity of second mass, with which it strikes the ground, is (take g=10 m/s2)a)3 m/sb)4 m/sc)5 m/sd)8 m/sCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A body at rest is dropped from a height h and strikes the ground with a velocity 3 m/s. Another body of same mass is dropped from the same height with an initial velocity of 4 m/s. The final velocity of second mass, with which it strikes the ground, is (take g=10 m/s2)a)3 m/sb)4 m/sc)5 m/sd)8 m/sCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body at rest is dropped from a height h and strikes the ground with a velocity 3 m/s. Another body of same mass is dropped from the same height with an initial velocity of 4 m/s. The final velocity of second mass, with which it strikes the ground, is (take g=10 m/s2)a)3 m/sb)4 m/sc)5 m/sd)8 m/sCorrect answer is option 'C'. Can you explain this answer?.
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