A body is released form the top of a tower of height h meters. It takes t seconds to reach the ground. Where is the ball at the time t/2 sec ?

Options

A. at h/2 meter from the ground
B. at h/4 meter from the ground
C. depends upon the mass and volume of the ball
D. at 3h/4 meter from the ground.

Explanation

To solve the problem step by step, we will use the equations of motion under uniform acceleration.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body is released from a height \( h \) meters.
- It takes \( t \) seconds to reach the ground.
- We need to find the position of the body at \( t/2 \) seconds.

2. **Initial Conditions**:
- The initial velocity \( u = 0 \) (since the body is released).
- The acceleration \( a = g \) (acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \)).

3. **Using the Second Equation of Motion**:
- The second equation of motion states:
\[ s = ut + \frac{1}{2} a t^2 \]
- For the time \( t/2 \), the distance \( x \) traveled from the top of the tower can be expressed as:
\[ x = 0 \cdot \left( \frac{t}{2} \right) + \frac{1}{2} g \left( \frac{t}{2} \right)^2 \]
- Simplifying this gives:
\[ x = \frac{1}{2} g \cdot \frac{t^2}{4} = \frac{g t^2}{8} \]

4. **Finding the Total Distance Fallen in Time \( t \)**:
- When the body reaches the ground in \( t \) seconds, the total distance fallen \( h \) can be expressed as:
\[ h = ut + \frac{1}{2} g t^2 \]
- Since \( u = 0 \), this simplifies to:
\[ h = \frac{1}{2} g t^2 \]

5. **Relating \( h \) and \( x \)**:
- From the equation for \( h \), we can express \( g t^2 \) as:
\[ g t^2 = 2h \]
- Substituting this into the equation for \( x \):
\[ x = \frac{g t^2}{8} = \frac{2h}{8} = \frac{h}{4} \]

6. **Finding the Position of the Ball**:
- The distance from the top of the tower at \( t/2 \) seconds is \( x = \frac{h}{4} \).
- Therefore, the distance from the ground at \( t/2 \) seconds is:
\[ h - x = h - \frac{h}{4} = \frac{3h}{4} \]

### Final Answer:
At \( t/2 \) seconds, the ball is \( \frac{h}{4} \) meters from the top of the tower and \( \frac{3h}{4} \) meters from the ground.

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