A projectile moving vertically upwards with a velocity of 200 m/s breaks into two equal parts at the height of 490 m.one part starts moving vertically upwards with a velocity of 400 m/s.How much time after the break up wil the other part hit the ground?

Options

A. `sqrt(10) s`
B. `2sqrt(10) s`
C. `5 s`
D. `10 s`

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the Problem
A projectile is moving upwards with an initial velocity of 200 m/s and breaks into two equal parts at a height of 490 m. One part moves upwards with a velocity of 400 m/s. We need to find the time it takes for the other part to hit the ground after the breakup.

### Step 2: Determine the Velocity of the Other Part
Using the conservation of momentum, we can find the velocity of the second part after the breakup.

- Before the breakup, the momentum of the projectile is:
\[ p_{\text{initial}} = m \cdot 200 \]

- After the breakup, the momentum of the two parts is:
\[ p_{\text{final}} = \frac{m}{2} \cdot 400 + \frac{m}{2} \cdot v \]
where \( v \) is the velocity of the second part.

Setting the initial momentum equal to the final momentum:
\[ m \cdot 200 = \frac{m}{2} \cdot 400 + \frac{m}{2} \cdot v \]

Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ 200 = 200 + \frac{v}{2} \]

Rearranging gives:
\[ 0 = \frac{v}{2} \implies v = 0 \]

### Step 3: Analyze the Motion of the Second Part
The second part has a velocity of 0 m/s at the moment of breakup. It will fall freely under the influence of gravity.

### Step 4: Use the Equation of Motion
We can use the second equation of motion to find the time taken to hit the ground:
\[ h = ut + \frac{1}{2} g t^2 \]
Where:
- \( h = -490 \) m (height is negative as it falls down)
- \( u = 0 \) m/s (initial velocity of the second part)
- \( g = 9.8 \) m/s² (acceleration due to gravity)

Substituting the values:
\[ -490 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
This simplifies to:
\[ -490 = 4.9 t^2 \]

### Step 5: Solve for Time \( t \)
Rearranging gives:
\[ t^2 = \frac{-490}{-4.9} = 100 \]
Taking the square root:
\[ t = \sqrt{100} = 10 \text{ seconds} \]

### Conclusion
The time after the breakup for the other part to hit the ground is **10 seconds**.

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