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A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of
- a)3 s
- b)5 s
- c)7 s
- d)9 s
Correct answer is option 'B'. Can you explain this answer?
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A body falls freely from rest. It covers as much distance in the last ...
Concept: Kinematics of Free Fall
When an object is dropped from a certain height, it falls freely under the influence of gravity. The motion of the object is called free fall.
Some important equations of kinematics of free fall are:
1. v = u + gt
2. s = ut + (1/2)gt^2
3. v^2 = u^2 + 2gs
Where,
v = final velocity
u = initial velocity (0 in free fall)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken
s = distance covered
Solution:
Given, the body falls freely from rest.
Let the time taken by the body to fall be 't' seconds.
It is given that the distance covered in the last second of motion is equal to the distance covered in the first three seconds.
So, the distance covered in the last second = distance covered in the first three seconds
Using equation (2), we can write:
s_last = u(t-1) + (1/2)g(t-1)^2
s_first_three = u(3) + (1/2)g(3)^2
Since s_last = s_first_three,
u(t-1) + (1/2)g(t-1)^2 = u(3) + (1/2)g(3)^2
Solving this equation, we get:
t = 5 seconds
Therefore, the body has fallen for a time of 5 seconds.
Hence, the correct answer is option 'B'.
When an object is dropped from a certain height, it falls freely under the influence of gravity. The motion of the object is called free fall.
Some important equations of kinematics of free fall are:
1. v = u + gt
2. s = ut + (1/2)gt^2
3. v^2 = u^2 + 2gs
Where,
v = final velocity
u = initial velocity (0 in free fall)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken
s = distance covered
Solution:
Given, the body falls freely from rest.
Let the time taken by the body to fall be 't' seconds.
It is given that the distance covered in the last second of motion is equal to the distance covered in the first three seconds.
So, the distance covered in the last second = distance covered in the first three seconds
Using equation (2), we can write:
s_last = u(t-1) + (1/2)g(t-1)^2
s_first_three = u(3) + (1/2)g(3)^2
Since s_last = s_first_three,
u(t-1) + (1/2)g(t-1)^2 = u(3) + (1/2)g(3)^2
Solving this equation, we get:
t = 5 seconds
Therefore, the body has fallen for a time of 5 seconds.
Hence, the correct answer is option 'B'.
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A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time ofa)3 sb)5 sc)7 sd)9 sCorrect answer is option 'B'. Can you explain this answer?
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