Show that there are two values of time for which a projectile is at the same height. Also show mathematically that the sum of these two times is equal to the time of flight.

Explanation

To show that there are two values of time for which a projectile is at the same height and to demonstrate mathematically that the sum of these two times is equal to the time of flight, we can follow these steps:

### Step 1: Understand the Motion of the Projectile
When a projectile is launched at an angle \( \theta \) with an initial velocity \( u \), its motion can be analyzed in two dimensions: horizontal and vertical. The vertical motion is influenced by gravity.

### Step 2: Write the Equation for Vertical Displacement
The vertical displacement \( y \) of the projectile can be described by the equation:
\[ y = u \sin \theta \cdot t - \frac{1}{2} g t^2 \]
where:
- \( u \sin \theta \) is the initial vertical velocity,
- \( g \) is the acceleration due to gravity,
- \( t \) is the time.

### Step 3: Rearrange the Equation
Rearranging the equation gives us:
\[ \frac{g}{2} t^2 - u \sin \theta \cdot t + y = 0 \]
This is a quadratic equation in \( t \).

### Step 4: Identify the Quadratic Formula
For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the solutions for \( x \) can be found using:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, \( a = \frac{g}{2} \), \( b = -u \sin \theta \), and \( c = y \).

### Step 5: Calculate the Two Values of Time
Using the quadratic formula, we find the two times \( t_1 \) and \( t_2 \):
\[ t_{1,2} = \frac{u \sin \theta \pm \sqrt{(u \sin \theta)^2 - 2gy}}{g/2} \]
This gives us two distinct times \( t_1 \) and \( t_2 \) at which the projectile is at height \( y \).

### Step 6: Sum of the Two Times
Now, we need to show that the sum of these two times is equal to the time of flight \( T \). The time of flight for a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]
Adding \( t_1 \) and \( t_2 \):
\[ t_1 + t_2 = \frac{u \sin \theta + \sqrt{(u \sin \theta)^2 - 2gy}}{g/2} + \frac{u \sin \theta - \sqrt{(u \sin \theta)^2 - 2gy}}{g/2} \]
The square root terms cancel out:
\[ t_1 + t_2 = \frac{2u \sin \theta}{g/2} = \frac{2u \sin \theta}{g} = T \]

### Conclusion
Thus, we have shown that there are two values of time for which a projectile is at the same height, and the sum of these two times is equal to the time of flight.

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