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A body is projected vertically upward with speed 10 m/s and other at same time with same speed in downward direction from the top of a tower. The magnitude of acceleration of first body w.r.t. second is {take g = 10 m/s2} (1) Zero (2) 10 m/s2 (3) 5 m/s2 (4) 20 m/s2?
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? A body is projected vertically upward with speed 10 m/s and other at...
**Solution:**
To solve this problem, let's consider the motion of the two bodies separately and then calculate their relative acceleration.
First, let's analyze the motion of the body projected vertically upward.
**Motion of the Body Projected Upward:**
- Initial velocity (u) = 10 m/s (upward)
- Final velocity (v) = 0 m/s (at the highest point, when the body momentarily comes to rest)
- Acceleration (a) = ?
- Time taken (t) = ?
We can use the equation of motion: v = u + at, where v = 0 and u = 10 m/s.
0 = 10 + a × t
Rearranging the equation, we get:
a × t = -10
This equation tells us that the acceleration and time taken are related in such a way that their product is equal to -10.
Now, let's analyze the motion of the body projected downward.
**Motion of the Body Projected Downward:**
- Initial velocity (u') = -10 m/s (downward)
- Final velocity (v') = ?
- Acceleration (a') = ?
- Time taken (t') = ?
The acceleration due to gravity (g) acts downward on the body, which is equal to 10 m/s^2.
We can use the equation of motion: v' = u' + a' × t', where v' is the final velocity.
v' = -10 + a' × t'
Since the final velocity is unknown, we cannot directly solve for acceleration or time taken. However, we can relate the acceleration of the second body (a') to the acceleration of the first body (a) using the concept of relative acceleration.
**Relative Acceleration:**
The relative acceleration of the first body with respect to the second body is given by the difference in their accelerations.
Relative acceleration (a_rel) = a - a'
Substituting the value of a from the first equation, we get:
Relative acceleration (a_rel) = (-10/t) - a'
Comparing this equation with the equation of motion for the second body, we can see that the relative acceleration is equal to the negative of the acceleration due to gravity.
Therefore, the magnitude of the relative acceleration is:
|Relative acceleration| = |-10/t - a'| = |-10/(t') - g|
Since g = 10 m/s^2, we can rewrite the equation as:
|Relative acceleration| = |-10/(t') - 10|
Simplifying further:
|Relative acceleration| = |-10(t' + 1)/t'|
As we can see, the magnitude of the relative acceleration depends on the value of t', the time taken by the second body to reach the ground.
Depending on the value of t', the magnitude of the relative acceleration can be zero, 10 m/s^2, 5 m/s^2, or 20 m/s^2.
Therefore, the correct answer is (2) 10 m/s^2.
To solve this problem, let's consider the motion of the two bodies separately and then calculate their relative acceleration.
First, let's analyze the motion of the body projected vertically upward.
**Motion of the Body Projected Upward:**
- Initial velocity (u) = 10 m/s (upward)
- Final velocity (v) = 0 m/s (at the highest point, when the body momentarily comes to rest)
- Acceleration (a) = ?
- Time taken (t) = ?
We can use the equation of motion: v = u + at, where v = 0 and u = 10 m/s.
0 = 10 + a × t
Rearranging the equation, we get:
a × t = -10
This equation tells us that the acceleration and time taken are related in such a way that their product is equal to -10.
Now, let's analyze the motion of the body projected downward.
**Motion of the Body Projected Downward:**
- Initial velocity (u') = -10 m/s (downward)
- Final velocity (v') = ?
- Acceleration (a') = ?
- Time taken (t') = ?
The acceleration due to gravity (g) acts downward on the body, which is equal to 10 m/s^2.
We can use the equation of motion: v' = u' + a' × t', where v' is the final velocity.
v' = -10 + a' × t'
Since the final velocity is unknown, we cannot directly solve for acceleration or time taken. However, we can relate the acceleration of the second body (a') to the acceleration of the first body (a) using the concept of relative acceleration.
**Relative Acceleration:**
The relative acceleration of the first body with respect to the second body is given by the difference in their accelerations.
Relative acceleration (a_rel) = a - a'
Substituting the value of a from the first equation, we get:
Relative acceleration (a_rel) = (-10/t) - a'
Comparing this equation with the equation of motion for the second body, we can see that the relative acceleration is equal to the negative of the acceleration due to gravity.
Therefore, the magnitude of the relative acceleration is:
|Relative acceleration| = |-10/t - a'| = |-10/(t') - g|
Since g = 10 m/s^2, we can rewrite the equation as:
|Relative acceleration| = |-10/(t') - 10|
Simplifying further:
|Relative acceleration| = |-10(t' + 1)/t'|
As we can see, the magnitude of the relative acceleration depends on the value of t', the time taken by the second body to reach the ground.
Depending on the value of t', the magnitude of the relative acceleration can be zero, 10 m/s^2, 5 m/s^2, or 20 m/s^2.
Therefore, the correct answer is (2) 10 m/s^2.
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? A body is projected vertically upward with speed 10 m/s and other at same time with same speed in downward direction from the top of a tower. The magnitude of acceleration of first body w.r.t. second is {take g = 10 m/s2} (1) Zero (2) 10 m/s2 (3) 5 m/s2 (4) 20 m/s2?
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