Two projectiles are thrown from the same point simultaneously with same velocity `10 ms^(-1)`. One goes straight vertically while other at `60^(@)` with the vertical . What will be the distance of separation between the after 1 second of their throw?

Options

A. `20 m`
B. `10 m`
C. `5 m`
D. `15 m`

Explanation

To solve the problem of finding the distance of separation between two projectiles thrown simultaneously, we can break it down into the following steps:

### Step 1: Determine the components of the velocities
- For the projectile thrown vertically (let's call it Projectile A), its velocity is entirely in the vertical direction:
- \( v_{A_y} = 10 \, \text{m/s} \)
- \( v_{A_x} = 0 \, \text{m/s} \)

- For the projectile thrown at an angle of \( 60^\circ \) with the vertical (let's call it Projectile B), we need to resolve its velocity into horizontal and vertical components:
- The vertical component \( v_{B_y} = 10 \cos(60^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s} \)
- The horizontal component \( v_{B_x} = 10 \sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \)

### Step 2: Calculate the positions of both projectiles after 1 second
- For Projectile A (vertical):
- Vertical position after 1 second:
\[ y_A = v_{A_y} \cdot t = 10 \cdot 1 = 10 \, \text{m} \]
- Horizontal position:
\[ x_A = 0 \, \text{m} \]

- For Projectile B (at \( 60^\circ \)):
- Vertical position after 1 second:
\[ y_B = v_{B_y} \cdot t = 5 \cdot 1 = 5 \, \text{m} \]
- Horizontal position after 1 second:
\[ x_B = v_{B_x} \cdot t = 5\sqrt{3} \cdot 1 = 5\sqrt{3} \, \text{m} \]

### Step 3: Calculate the distance of separation
- The coordinates of Projectile A after 1 second are \( (0, 10) \).
- The coordinates of Projectile B after 1 second are \( (5\sqrt{3}, 5) \).

- The distance \( d \) between the two projectiles can be calculated using the distance formula:
\[ d = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \]
Substituting the values:
\[ d = \sqrt{(5\sqrt{3} - 0)^2 + (5 - 10)^2} \]
\[ d = \sqrt{(5\sqrt{3})^2 + (-5)^2} \]
\[ d = \sqrt{75 + 25} = \sqrt{100} = 10 \, \text{m} \]

### Conclusion
The distance of separation between the two projectiles after 1 second is \( 10 \, \text{m} \).

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