A cricket ball thrown across a field is a heights `h_(1)` and `h_(2)` from the point of projection at time `t_(1)` and `t_(2)` respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is

Options

A. `((h_(1)t_(2)^(2)-h_(2)t_(1)^(2))/(h_(1)t_(2) -h_(2)t_(1)))`
B. `((h_(1)t_(2)^(2)-h_(2)t_(1)^(2))/(h_(1)t_(2)-h_(2)t_(1)))`
C. `((h_(1)t_(2)^(2)-h_(2)t_(1)^(2))/(h_(1)t_(2)-h_(2)t_(1)))`
D. None of these

Explanation

To find the time of flight of a cricket ball thrown across a field and caught at the same height from which it was thrown, we can use the principles of projectile motion. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a cricket ball thrown at an angle with an initial velocity \( u \). It reaches heights \( h_1 \) and \( h_2 \) at times \( t_1 \) and \( t_2 \) respectively. The ball is caught at the same height from which it was thrown.

### Step 2: Write the Equations for Vertical Motion
The vertical motion of the ball can be described using the following kinematic equation:
\[ h = u_y t - \frac{1}{2} g t^2 \]
where \( u_y = u \sin \theta \) is the vertical component of the initial velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time.

For the heights \( h_1 \) and \( h_2 \):
1. At time \( t_1 \):
\[ h_1 = u \sin \theta \cdot t_1 - \frac{1}{2} g t_1^2 \]
2. At time \( t_2 \):
\[ h_2 = u \sin \theta \cdot t_2 - \frac{1}{2} g t_2^2 \]

### Step 3: Rearranging the Equations
We can rearrange both equations to isolate \( u \sin \theta \):
1. From the first equation:
\[ u \sin \theta = \frac{h_1 + \frac{1}{2} g t_1^2}{t_1} \]
2. From the second equation:
\[ u \sin \theta = \frac{h_2 + \frac{1}{2} g t_2^2}{t_2} \]

### Step 4: Set the Two Expressions Equal
Since both expressions equal \( u \sin \theta \), we can set them equal to each other:
\[ \frac{h_1 + \frac{1}{2} g t_1^2}{t_1} = \frac{h_2 + \frac{1}{2} g t_2^2}{t_2} \]

### Step 5: Cross-Multiply and Simplify
Cross-multiplying gives:
\[ (h_1 + \frac{1}{2} g t_1^2) t_2 = (h_2 + \frac{1}{2} g t_2^2) t_1 \]
Expanding both sides leads to:
\[ h_1 t_2 + \frac{1}{2} g t_1^2 t_2 = h_2 t_1 + \frac{1}{2} g t_2^2 t_1 \]

### Step 6: Rearranging for Time of Flight
We know that the total time of flight \( T \) for a projectile is given by:
\[ T = \frac{2 u \sin \theta}{g} \]
Using the expressions for \( u \sin \theta \) derived earlier, we can substitute back to find \( T \).

### Step 7: Final Expression for Time of Flight
After performing the necessary algebraic manipulations, we arrive at the final expression for the total time of flight:
\[ T = \frac{h_1 t_2^2 - h_2 t_1^2}{h_1 t_2 - h_2 t_1} \]

### Conclusion
Thus, the time of flight of the ball is:
\[ T = \frac{h_1 t_2^2 - h_2 t_1^2}{h_1 t_2 - h_2 t_1} \]