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)m//s`
B. `2sqrt(5)m//s`
C. `5sqrt(4)m//s`
D. `10m//s`

Explanation

To solve the problem, we need to determine the velocity of the center of mass of the cracker when it reaches a height of 1 meter from the ground. Here’s a step-by-step solution:

### Step 1: Analyze the Initial Conditions
The cracker is thrown with an initial velocity of \(10 \, \text{m/s}\) at an angle of \(45^\circ\) with the vertical. We can resolve this velocity into its horizontal and vertical components.

- **Vertical Component (\(v_{y0}\))**:
\[ v_{y0} = 10 \cos(45^\circ) = 10 \cdot \frac{1}{\sqrt{2}} \approx 7.07 \, \text{m/s} \]

- **Horizontal Component (\(v_{x0}\))**:
\[ v_{x0} = 10 \sin(45^\circ) = 10 \cdot \frac{1}{\sqrt{2}} \approx 7.07 \, \text{m/s} \]

### Step 2: Determine the Time to Reach 1 Meter Height
We can use the kinematic equation for vertical motion to find the time it takes to reach a height of 1 meter. The equation is:
\[ h = v_{y0} t - \frac{1}{2} g t^2 \]
where \(h = 1 \, \text{m}\) and \(g = 10 \, \text{m/s}^2\).

Substituting the values:
\[ 1 = 7.07 t - \frac{1}{2} \cdot 10 t^2 \]
\[ 1 = 7.07 t - 5 t^2 \]
Rearranging gives us:
\[ 5 t^2 - 7.07 t + 1 = 0 \]

### Step 3: Solve the Quadratic Equation
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- Here, \(a = 5\), \(b = -7.07\), and \(c = 1\).
\[ t = \frac{7.07 \pm \sqrt{(-7.07)^2 - 4 \cdot 5 \cdot 1}}{2 \cdot 5} \]
\[ t = \frac{7.07 \pm \sqrt{50.0049 - 20}}{10} \]
\[ t = \frac{7.07 \pm \sqrt{30.0049}}{10} \]
Calculating the square root:
\[ \sqrt{30.0049} \approx 5.48 \]
So,
\[ t = \frac{7.07 \pm 5.48}{10} \]
Calculating the two possible times:
\[ t_1 = \frac{12.55}{10} \approx 1.255 \, \text{s} \quad \text{(not relevant as it exceeds time to reach 1m)} \]
\[ t_2 = \frac{1.59}{10} \approx 0.159 \, \text{s} \]

### Step 4: Calculate the Velocity of the Center of Mass
Since the explosion does not affect the center of mass due to internal forces, the center of mass continues to move with the same velocity it had just before the explosion.

The vertical velocity at height \(1 \, \text{m}\) can be calculated using:
\[ v_y = v_{y0} - g t \]
Substituting \(t = 0.159 \, \text{s}\):
\[ v_y = 7.07 - 10 \cdot 0.159 \approx 7.07 - 1.59 \approx 5.48 \, \text{m/s} \]

### Step 5: Combine the Velocity Components
The total velocity of the center of mass is given by combining the horizontal and vertical components:
\[ v_{cm} = \sqrt{v_x^2 + v_y^2} \]
Substituting the values:
\[ v_{cm} = \sqrt{(7.07)^2 + (5.48)^2} \]
Calculating:
\[ v_{cm} = \sqrt{50 + 30.00} \approx \sqrt{80} \approx 8.94 \, \text{m/s} \]

### Final Answer
The velocity of the center of mass when it is at a height of \(1 \, \text{m}\) from the ground is approximately \(8.94 \, \text{m/s}\).

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