A body is projected upwards with a velocity `u`. It passes through a certain point above the ground after `t_(1)`, Find the time after which the body passes through the same point during the journey.

Options

A. `((u)/(g)-t_(1)^(2))`
B. `2(u)/(g)-t_(1)`
C. `3((u^(2))/(g)-t_(1))`
D. `3((u^(2))/(g^(2))-t_(1))`

Answer

To solve the problem step by step, we will use the equations of motion under uniform acceleration. Here’s how we can derive the time after which the body passes through the same point during its journey:

### Step 1: Understand the Motion
A body is projected upwards with an initial velocity \( u \). As it rises, it will eventually reach a maximum height and then start descending. The question states that it passes through a certain height \( h \) at time \( t_1 \) during its upward journey.

### Step 2: Write the Equation of Motion
We can use the second equation of motion to relate the height \( h \) to the time \( t \):
\[ h = ut - \frac{1}{2}gt^2 \]
where:
- \( h \) is the height,
- \( u \) is the initial velocity,
- \( g \) is the acceleration due to gravity (acting downwards),
- \( t \) is the time.

### Step 3: Set Up the Equation for Height \( h \)
At time \( t_1 \), the body reaches height \( h \):
\[ h = ut_1 - \frac{1}{2}gt_1^2 \]
Now, we can rearrange this equation:
\[ \frac{1}{2}gt_1^2 - ut_1 + h = 0 \]

### Step 4: Recognize the Quadratic Equation
This is a quadratic equation in \( t \):
\[ \frac{1}{2}g t^2 - ut + h = 0 \]
The general form of a quadratic equation is \( at^2 + bt + c = 0 \), where:
- \( a = \frac{1}{2}g \),
- \( b = -u \),
- \( c = h \).

### Step 5: Use the Quadratic Formula
The roots of the quadratic equation can be found using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting our values:
\[ t = \frac{u \pm \sqrt{u^2 - 2gh}}{g} \]

### Step 6: Identify the Roots
Since the body passes through the height \( h \) twice (once while going up and once while coming down), we denote the two times as \( t_1 \) and \( t_2 \). The two roots are:
- \( t_1 \) (upward journey),
- \( t_2 \) (downward journey).

### Step 7: Sum of the Roots
According to the properties of quadratic equations, the sum of the roots \( t_1 + t_2 \) can be given by:
\[ t_1 + t_2 = -\frac{b}{a} = \frac{2u}{g} \]

### Step 8: Solve for \( t_2 \)
To find \( t_2 \):
\[ t_2 = \frac{2u}{g} - t_1 \]

### Final Answer
Thus, the time after which the body passes through the same point during its journey is:
\[ t_2 = \frac{2u}{g} - t_1 \]