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A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three second of its motion.the stone remains in the air of?
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A stone falls freely from rest and the total distance covered by it in...
The distance travelled by the stone in the nth second is given by,

5th second is the last second. So, the stone will remain in air for 5 seconds.
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A stone falls freely from rest and the total distance covered by it in...
Understanding the Scenario


To solve this problem, let's first understand the scenario. We have a stone that falls freely from rest. This means that the stone is only under the influence of gravity, and there are no other forces acting on it. We need to determine how long the stone remains in the air.

Using Equations of Motion


We can use the equations of motion to solve this problem. The equation that relates distance, time, and acceleration for an object in free fall is:

s = ut + (1/2)at^2

Where:
- s is the distance covered
- u is the initial velocity (which is zero as the stone starts from rest)
- t is the time taken
- a is the acceleration due to gravity (which is approximately 9.8 m/s^2)

Calculating the Distance


Let's break down the problem into two parts: the first three seconds and the last second.

First Three Seconds:
Using the equation of motion, we can calculate the distance covered by the stone in the first three seconds:

s = 0 + (1/2)(9.8)(3)^2
s = 0 + (1/2)(9.8)(9)
s = 0 + 44.1
s = 44.1 meters

Last Second:
Now, we know that the distance covered in the last second is equal to the distance covered in the first three seconds. Let's denote this distance as 'd'.

d = 44.1 meters

Calculating the Time


We can rearrange the equation of motion to solve for time:

s = ut + (1/2)at^2
t^2 + 2(s/a)t - 2(0) = 0

Let's substitute the values:
t^2 + (2d/a)t = 0

Now, we can solve this quadratic equation using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

Where:
- a = 1
- b = 2d/a = 2d/9.8
- c = 0

t = (-2d/9.8 ± √((2d/9.8)^2 - 4(1)(0))) / 2(1)
t = (-2d/9.8 ± √((4d^2/9.8^2))) / 2
t = (-2d/9.8 ± (2d/9.8)) / 2
t = -d/9.8 + d/9.8
t = 0

Conclusion


From our calculation, we find that the time taken for the stone to cover the distance 'd' in the last second is zero. This means that the stone remains in the air for zero seconds in the last second of its motion. Therefore
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A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three second of its motion.the stone remains in the air of?
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A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three second of its motion.the stone remains in the air of? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three second of its motion.the stone remains in the air of? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three second of its motion.the stone remains in the air of?.
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