A body is projected vertically upwatds at time t=0 and is it seen at a height H at time `t_1` and `t_2` second during its flight. The maximum height attainet is (g is acceleration due to garavity).

Options

A. `g/4 (t_1+t_2)^2`
B. `g ((t_1+t_2)/4)^2`
C. `2g((t_1+t_2)/4)^2`
D. `g/4(t_1t_2)`

Answer

To find the maximum height attained by a body projected vertically upwards, given that it is seen at a height \( H \) at times \( t_1 \) and \( t_2 \), we can follow these steps:

### Step 1: Understand the motion of the body
When a body is projected upwards, it will rise to a maximum height and then fall back down. The times \( t_1 \) and \( t_2 \) indicate two instances when the body is at the same height \( H \) during its ascent and descent.

### Step 2: Relate the times \( t_1 \) and \( t_2 \)
The time taken to go from height \( H \) to the maximum height is equal to the time taken to fall from the maximum height back down to height \( H \). Therefore, the average time taken to reach the maximum height after reaching \( H \) can be expressed as:
\[ t_{\text{avg}} = \frac{t_1 + t_2}{2} \]

### Step 3: Determine the total time of flight
The total time of flight \( T \) can be expressed as:
\[ T = t_2 + (t_2 - t_1) = t_1 + t_2 \]

### Step 4: Use the kinematic equation for height
The height \( H \) at time \( t \) can be expressed using the kinematic equation:
\[ H = ut - \frac{1}{2} g t^2 \]
where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity.

### Step 5: Substitute the average time into the height equation
Substituting \( t = \frac{t_1 + t_2}{2} \) into the height equation gives:
\[ H = u \left(\frac{t_1 + t_2}{2}\right) - \frac{1}{2} g \left(\frac{t_1 + t_2}{2}\right)^2 \]

### Step 6: Rearranging for maximum height
The maximum height \( H_{\text{max}} \) can be derived from the time of flight. The time to reach maximum height \( t_{\text{max}} \) is equal to \( \frac{T}{2} = \frac{t_1 + t_2}{2} \). The maximum height can be expressed as:
\[ H_{\text{max}} = u \cdot t_{\text{max}} - \frac{1}{2} g t_{\text{max}}^2 \]

### Step 7: Substitute \( t_{\text{max}} \)
Substituting \( t_{\text{max}} = \frac{t_1 + t_2}{2} \):
\[ H_{\text{max}} = u \cdot \frac{t_1 + t_2}{2} - \frac{1}{2} g \left(\frac{t_1 + t_2}{2}\right)^2 \]

### Step 8: Simplify the expression
This can be simplified to:
\[ H_{\text{max}} = \frac{(t_1 + t_2)u}{2} - \frac{g(t_1 + t_2)^2}{8} \]

### Step 9: Find the relationship between \( H \) and \( H_{\text{max}} \)
From the earlier equations, we can derive:
\[ H = \frac{g}{8} (t_1 + t_2)^2 \]
Thus, the maximum height attained can be expressed as:
\[ H_{\text{max}} = \frac{g}{8} (t_1 + t_2)^2 \]

### Conclusion
The maximum height attained by the body is:
\[ H_{\text{max}} = \frac{g}{8} (t_1 + t_2)^2 \]