An object projected upward acquires a velocity of 9.8 `ms^(-1)`, when it reaches half of the maximum height, The maximum height reached is :-

Options

A. 4.9 m
B. 7.8 m
C. 8.8 m
D. 9.8 m

Answer

To solve the problem, we need to determine the maximum height reached by an object projected upward, given that it has a velocity of 9.8 m/s when it reaches half of that maximum height.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We are given that the object reaches a velocity of 9.8 m/s at half of the maximum height (h/2).
- We need to find the maximum height (h).

2. **Using the Equation of Motion**:
- We will use the equation of motion:
\[ v^2 = u^2 + 2as \]
- Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration (which will be negative due to gravity), and \(s\) is the distance traveled.

3. **Setting Up the Equation**:
- At half the maximum height (h/2), the velocity \(v = 9.8 \, \text{m/s}\).
- The acceleration due to gravity \(a = -g = -9.8 \, \text{m/s}^2\) (acting downwards).
- The distance \(s = h/2\).
- We can rewrite the equation as:
\[ (9.8)^2 = u^2 - 2g \left(\frac{h}{2}\right) \]

4. **Finding the Initial Velocity (u)**:
- Rearranging the equation gives:
\[ u^2 = (9.8)^2 + g \cdot h \]
- We know \(g = 9.8 \, \text{m/s}^2\), so substituting this in:
\[ u^2 = (9.8)^2 + 9.8 \cdot h \]

5. **Using the Maximum Height Formula**:
- The maximum height \(h\) can also be expressed using the initial velocity \(u\):
\[ h = \frac{u^2}{2g} \]
- Substituting \(u^2\) from the previous step into this equation:
\[ h = \frac{(9.8)^2 + 9.8h}{2g} \]

6. **Solving for h**:
- Rearranging gives:
\[ 2gh = (9.8)^2 + 9.8h \]
- Bringing all terms involving \(h\) to one side:
\[ 2gh - 9.8h = (9.8)^2 \]
- Factoring out \(h\):
\[ h(2g - 9.8) = (9.8)^2 \]
- Solving for \(h\):
\[ h = \frac{(9.8)^2}{2g - 9.8} \]
- Substituting \(g = 9.8 \, \text{m/s}^2\):
\[ h = \frac{(9.8)^2}{9.8} = 9.8 \, \text{m} \]

### Final Answer:
The maximum height reached by the object is **9.8 meters**.