Two particles are simultaneously thrown in horizontal direction from two points on a riverbank, which are at certain height above the water surface. The initial velocities of the particles are `v_1= 5m//s and v_2 = 7.5 m//s` respectively. Both particles fall into the water at the same time. First particles enters the water at a point s = 10 m from the bank. Determine
(a) the time of flight of the two particles,
(b) the height from which they are thrown,
(c) the point where the second particle falls in water.

Explanation

To solve the problem step by step, we will analyze the motion of both particles thrown horizontally from the same height.

### Step 1: Determine the time of flight of the two particles

Both particles are thrown simultaneously and fall into the water at the same time. The time of flight can be determined using the horizontal motion of the first particle.

Given:
- Horizontal distance traveled by the first particle, \( s_1 = 10 \, \text{m} \)
- Initial velocity of the first particle, \( v_1 = 5 \, \text{m/s} \)

Using the formula for horizontal motion:
\[ s = v \cdot t \]
We can rearrange this to find the time \( t \):
\[ t = \frac{s_1}{v_1} = \frac{10 \, \text{m}}{5 \, \text{m/s}} = 2 \, \text{s} \]

### Step 2: Determine the height from which they are thrown

Now that we have the time of flight, we can calculate the height \( H \) from which both particles are thrown. The vertical motion can be described by the equation:
\[ H = \frac{1}{2} g t^2 \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

Substituting the values:
\[ H = \frac{1}{2} \cdot 10 \, \text{m/s}^2 \cdot (2 \, \text{s})^2 = \frac{1}{2} \cdot 10 \cdot 4 = 20 \, \text{m} \]

### Step 3: Determine the point where the second particle falls in water

For the second particle, we need to calculate the horizontal distance it travels before hitting the water. The initial velocity of the second particle is \( v_2 = 7.5 \, \text{m/s} \). Using the same time of flight \( t = 2 \, \text{s} \):
\[ s_2 = v_2 \cdot t = 7.5 \, \text{m/s} \cdot 2 \, \text{s} = 15 \, \text{m} \]

### Summary of Results
(a) Time of flight of the two particles: \( t = 2 \, \text{s} \)
(b) Height from which they are thrown: \( H = 20 \, \text{m} \)
(c) The point where the second particle falls in water: \( s_2 = 15 \, \text{m} \)

---