A body is thrown vertically up to reach its maximum height in t seconds. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is:

Options

A. `sqrt2 t`
B. `(1 + (1)/(sqrt2))t`
C. `(3t)/(2)`
D. `(t)/(sqrt2)`

Explanation

To solve the problem, we need to find the total time taken for a body thrown vertically upward to reach half of its maximum height while returning. Let's break this down step by step.

### Step 1: Understand the motion
When a body is thrown vertically upwards, it takes time \( t \) to reach its maximum height. At maximum height, the velocity becomes zero. The time taken to reach maximum height is equal to the time taken to return from maximum height to the ground.

### Step 2: Determine maximum height
Let the maximum height reached by the body be \( h \). The formula for the maximum height in terms of time \( t \) is given by:
\[ h = \frac{1}{2} g t^2 \]
where \( g \) is the acceleration due to gravity.

### Step 3: Find the time to reach half of the maximum height
Half of the maximum height is \( \frac{h}{2} \). We need to find the time taken to reach this height while coming down.

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s = \frac{h}{2} \)
- \( u = 0 \) (initial velocity at the maximum height)
- \( a = g \) (acceleration due to gravity)

Substituting these values, we get:
\[ \frac{h}{2} = 0 + \frac{1}{2} g t_1^2 \]
This simplifies to:
\[ \frac{h}{2} = \frac{1}{2} g t_1^2 \implies h = g t_1^2 \]
Now, substituting \( h = \frac{1}{2} g t^2 \) from Step 2 into this equation:
\[ \frac{1}{2} g t^2 = g t_1^2 \]
Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ \frac{1}{2} t^2 = t_1^2 \]
Taking the square root of both sides gives:
\[ t_1 = \frac{t}{\sqrt{2}} \]

### Step 4: Total time from projection to half maximum height while returning
The total time taken to reach half of the maximum height while returning consists of two parts:
1. The time taken to go up to the maximum height \( t \).
2. The time taken to come down to half the maximum height \( t_1 = \frac{t}{\sqrt{2}} \).

Thus, the total time \( T \) is:
\[ T = t + t_1 = t + \frac{t}{\sqrt{2}} \]
Factoring out \( t \):
\[ T = t \left( 1 + \frac{1}{\sqrt{2}} \right) \]

### Final Answer
The total time from the time of projection to reach a point at half of its maximum height while returning is:
\[ T = t \left( 1 + \frac{1}{\sqrt{2}} \right) \]