Question

If a body released from the top of a tower of height H meter takes T seconds to reach the ground, where is the body at time T/2 seconds from the ground?

• A. \(\frac{H}{2}\)
• B. \(\frac{H}{4}\)
• C. \(\frac{3H}{4}\)
• D. \(\frac{2H}{3}\)

Hint:

For an object in free fall from rest, the distance covered is proportional to the square of the time (\(s \propto t^2\)). This means in half the total time, it covers \((1/2)^2 = 1/4\) of the total distance. Therefore, the remaining distance to the ground is \(1 - 1/4 = 3/4\) of the total height.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A body falls from rest from a height H, taking time T. We need to find its height from the ground at time T/2.

Step 2: Key Formula or Approach:
We use the equation of motion for distance traveled under constant acceleration, starting from rest:
\[ s = ut + \frac{1}{2}at^2 \]
Since the body is released from rest, \(u=0\). Let's take the downward direction as positive, so \(a=g\). The distance fallen from the top is \(s = \frac{1}{2}gt^2\).

Step 3: Detailed Explanation:
First, relate the total height H to the total time T. In time T, the body falls a distance H.
\[ H = \frac{1}{2}gT^2 \quad \text{(Equation 1)} \]
Next, find the distance the body has fallen from the top at time \(t = T/2\). Let's call this distance \(s_{T/2}\).
\[ s_{T/2} = \frac{1}{2}g\left(\frac{T}{2}\right)^2 = \frac{1}{2}g\frac{T^2}{4} = \frac{1}{4} \left(\frac{1}{2}gT^2\right) \]
From Equation 1, we know that \(\frac{1}{2}gT^2 = H\). So, we can substitute H into the expression for \(s_{T/2}\):
\[ s_{T/2} = \frac{H}{4} \]
This is the distance fallen from the top of the tower. The question asks for the position (height) of the body from the ground.
\[ \text{Height from ground} = \text{Total Height} - \text{Distance fallen} \]
\[ \text{Height from ground} = H - s_{T/2} = H - \frac{H}{4} = \frac{3H}{4} \]

Step 4: Final Answer:
At time T/2, the body is at a height of \(\frac{3H}{4}\) from the ground.