A cracker is thrown into air with a velocity of `10 m//s` at an angle of `45^(@)` with the vertical. When it is at a height of `0.5 m` from the ground, it explodes into a number of pieces which follow different parabolic paths. What is the velocity of centre of mass, when it is at a height of `1 m` from the ground? (`g = 10 m//s^(2)`)

Options

A. `4sqrt(5)ms^(-1)`
B. `2sqrt(5)ms^(-1)`
C. `5sqrt(4)ms^(-1)`
D. `5ms^(-1)`

Explanation

To solve the problem step by step, we will analyze the motion of the cracker before and after it explodes, focusing on the center of mass.

### Step 1: Determine the initial velocity components
The cracker is thrown with an initial velocity \( u = 10 \, \text{m/s} \) at an angle of \( 45^\circ \) with the vertical. We can find the horizontal (\( u_x \)) and vertical (\( u_y \)) components of the initial velocity.

- Since the angle is with respect to the vertical, we have:
\[ u_x = u \sin(45^\circ) = 10 \cdot \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, \text{m/s} \]
\[ u_y = u \cos(45^\circ) = 10 \cdot \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, \text{m/s} \]

### Step 2: Analyze the motion at the height of 1 meter
The cracker explodes at a height of \( 0.5 \, \text{m} \) and we need to find the velocity of the center of mass when it reaches a height of \( 1 \, \text{m} \).

### Step 3: Calculate the vertical velocity at height 1 meter
To find the vertical velocity (\( v_y \)) at height \( 1 \, \text{m} \), we can use the kinematic equation:
\[ v_y^2 = u_y^2 - 2g s \]
where:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( s = 1 \, \text{m} - 0.5 \, \text{m} = 0.5 \, \text{m} \) (displacement)

Substituting the values:
\[ v_y^2 = \left(\frac{10}{\sqrt{2}}\right)^2 - 2 \cdot 10 \cdot 0.5 \]
\[ v_y^2 = \frac{100}{2} - 10 = 50 - 10 = 40 \]
\[ v_y = \sqrt{40} = 2\sqrt{10} \, \text{m/s} \]

### Step 4: Determine the horizontal velocity
The horizontal velocity (\( v_x \)) remains constant since there are no horizontal forces acting on the cracker:
\[ v_x = u_x = \frac{10}{\sqrt{2}} \, \text{m/s} \]

### Step 5: Calculate the magnitude of the velocity of the center of mass
Now, we can find the magnitude of the velocity of the center of mass using the Pythagorean theorem:
\[ v_{\text{cm}} = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v_{\text{cm}} = \sqrt{\left(\frac{10}{\sqrt{2}}\right)^2 + (2\sqrt{10})^2} \]
\[ = \sqrt{50 + 40} = \sqrt{90} = 3\sqrt{10} \, \text{m/s} \]

### Final Answer
The velocity of the center of mass when it is at a height of \( 1 \, \text{m} \) from the ground is \( 3\sqrt{10} \, \text{m/s} \).

---