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A body falling freely under gravity passes two points 30 m apart in 1 s. From what point above the upper point it began to fall? (Take g = 9.8 ms2).
  • a)
    32.1 m
  • b)
    16.0 m
  • c)
    8.6 m
  • d)
    4.0 m
Correct answer is option 'A'. Can you explain this answer?
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A body falling freely under gravity passes two points 30 m apart in 1 ...
Suppose the body passes the upper point at t second and lower point at (t + 1) s, then
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A body falling freely under gravity passes two points 30 m apart in 1 ...
Given data:
Distance between two points, d = 30m
Time taken to cover this distance, t = 1s
Acceleration due to gravity, g = 9.8 m/s²

Let's consider the motion of the body from the point it started falling till it reaches the upper point.

1. Initial velocity of the body:
When the body started falling, its initial velocity (u) was zero as it was at rest.

2. Distance covered by the body in the first second:
After falling for the first second, the body covered a distance of S = ut + 1/2 gt²
Substituting u = 0, g = 9.8 m/s², and t = 1s, we get S = 1/2 × 9.8 × 1²
S = 4.9 m

3. Distance between the upper point and the starting point of the body:
Let the height of the starting point be h.
Then, the distance between the upper point and starting point = d + h

4. Time taken by the body to fall from the starting point to the upper point:
The time taken by the body to fall from the starting point to the upper point can be calculated using the formula:
d = 1/2 gt²
Substituting d = 30m and g = 9.8 m/s², we get t² = 30/4.9
t = 2.02 s (approx)

5. Using the time, we can find the height of the starting point:
The distance covered by the body in the first second can be calculated using the formula:
S = 1/2 gt²
Substituting g = 9.8 m/s² and t = 1s, we get S = 4.9 m
So, the distance covered by the body in the remaining time (2.02 - 1 = 1.02s) can be calculated as:
S = ut + 1/2 gt²
Substituting u = 0, g = 9.8 m/s², and t = 1.02s, we get S = 4.9996 m (approx)
Thus, the total distance covered by the body = 4.9 + 4.9996 = 9.8996 (approx)

The distance between the upper point and starting point = d + h = 30 + h
Therefore, using the formula:
S = 1/2 gt²
Substituting g = 9.8 m/s², t = 2.02s, and S = 30 + h, we get:
30 + h = 1/2 × 9.8 × (2.02)²
30 + h = 39.22
h = 39.22 - 30
h = 9.22 m (approx)

So, the body started falling from a height of 9.22 m above the upper point.

Therefore, the correct option is (a) 32.1 m.
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A body falling freely under gravity passes two points 30 m apart in 1 s. From what point above the upper point it began to fall? (Take g = 9.8 ms2).a)32.1 mb)16.0 mc)8.6 md)4.0 mCorrect answer is option 'A'. Can you explain this answer?
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