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A body is released from the top of the tower H metre high. It takes t seconds to reach the ground. Where is the body after t/2 seconds of release?
- a)At 3H/4 metres from the ground
- b)At H/2 metres from the ground
- c)At H/6 metres from the ground
- d)At H/4 metres from the ground
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A body is released from the top of the tower H metre high. It takes t...
Solution:
Given:
- A body is released from the top of a tower H meters high.
- It takes t seconds to reach the ground.
To find:
- Where is the body after t/2 seconds of release?
Explanation:
To solve this problem, let's consider the motion of the body using the equations of motion.
1. Initial velocity (u) = 0 m/s (as the body is released from rest)
2. Final velocity (v) = ?
3. Acceleration (a) = acceleration due to gravity = g = 9.8 m/s^2 (assuming no air resistance)
4. Time (t) = t seconds
5. Displacement (s) = ?
Using the equation of motion:
v = u + at
Since the initial velocity (u) is zero, the equation becomes:
v = at
The final velocity (v) of the body after t seconds is given by the product of acceleration (a) and time (t).
Now, let's consider the motion of the body after t/2 seconds.
1. Initial velocity (u) = 0 m/s
2. Final velocity (v) = ?
3. Acceleration (a) = g = 9.8 m/s^2
4. Time (t) = t/2 seconds (half the time to reach the ground)
5. Displacement (s) = ?
Using the equation of motion:
v = u + at
Substituting the given values:
v = 0 + g(t/2)
v = gt/2
The final velocity (v) of the body after t/2 seconds is given by the product of acceleration (g) and half the time (t/2).
Now, to find the displacement (s) of the body after t/2 seconds, we can use the equation of motion:
s = ut + (1/2)at^2
In this case, initial velocity (u) is zero, so the equation becomes:
s = (1/2)at^2
Substituting the given values:
s = (1/2)g(t/2)^2
s = (1/2)g(t^2/4)
s = (1/8)gt^2
The displacement (s) of the body after t/2 seconds is given by one-eighth of the product of acceleration (g) and the square of time (t).
From the above equation, we can see that the displacement (s) is directly proportional to the square of time (t), which means that the displacement increases as the square of time increases.
Therefore, after t/2 seconds of release, the body will be at a point that is closer to the ground compared to the initial position.
Option A) At 3H/4 meters from the ground is the correct answer.
Given:
- A body is released from the top of a tower H meters high.
- It takes t seconds to reach the ground.
To find:
- Where is the body after t/2 seconds of release?
Explanation:
To solve this problem, let's consider the motion of the body using the equations of motion.
1. Initial velocity (u) = 0 m/s (as the body is released from rest)
2. Final velocity (v) = ?
3. Acceleration (a) = acceleration due to gravity = g = 9.8 m/s^2 (assuming no air resistance)
4. Time (t) = t seconds
5. Displacement (s) = ?
Using the equation of motion:
v = u + at
Since the initial velocity (u) is zero, the equation becomes:
v = at
The final velocity (v) of the body after t seconds is given by the product of acceleration (a) and time (t).
Now, let's consider the motion of the body after t/2 seconds.
1. Initial velocity (u) = 0 m/s
2. Final velocity (v) = ?
3. Acceleration (a) = g = 9.8 m/s^2
4. Time (t) = t/2 seconds (half the time to reach the ground)
5. Displacement (s) = ?
Using the equation of motion:
v = u + at
Substituting the given values:
v = 0 + g(t/2)
v = gt/2
The final velocity (v) of the body after t/2 seconds is given by the product of acceleration (g) and half the time (t/2).
Now, to find the displacement (s) of the body after t/2 seconds, we can use the equation of motion:
s = ut + (1/2)at^2
In this case, initial velocity (u) is zero, so the equation becomes:
s = (1/2)at^2
Substituting the given values:
s = (1/2)g(t/2)^2
s = (1/2)g(t^2/4)
s = (1/8)gt^2
The displacement (s) of the body after t/2 seconds is given by one-eighth of the product of acceleration (g) and the square of time (t).
From the above equation, we can see that the displacement (s) is directly proportional to the square of time (t), which means that the displacement increases as the square of time increases.
Therefore, after t/2 seconds of release, the body will be at a point that is closer to the ground compared to the initial position.
Option A) At 3H/4 meters from the ground is the correct answer.
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Community Answer
A body is released from the top of the tower H metre high. It takes t...
Since the body is released from rest, initial velocity u = 0.
From equation of motion, we have
Displacement of body at time t/2 from the top,
Hence, height of body from the ground = 
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A body is released from the top of the tower H metre high. It takes t seconds to reach the ground. Where is the body after t/2 seconds of release?a)At 3H/4 metres from the groundb)At H/2 metres from the groundc)At H/6 metres from the groundd)At H/4 metres from the groundCorrect answer is option 'A'. Can you explain this answer?
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