A projectile moving vertically upwards with a velocity of 200 `ms^(-1)` breaks into equal parts at a height of 490 m.One part starts moving vertically upwards with a velocity of 400 `ms^(-1)` .How much time it will take,after the break up with the other part to hit the ground?

Options

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

Explanation

To solve the problem, we need to determine the time it takes for the second part of the projectile (which moves upwards with a velocity of 400 m/s after the breakup) to hit the ground after the breakup at a height of 490 m.

### Step-by-Step Solution:

1. **Identify the initial conditions after the breakup**:
- The height at which the projectile breaks is \( h = 490 \, \text{m} \).
- The initial velocity of the second part moving upwards is \( u = 400 \, \text{m/s} \).
- The acceleration due to gravity is \( g = 9.8 \, \text{m/s}^2 \) (acting downwards).

2. **Use the second equation of motion**:
The second equation of motion is given by:
\[ h = ut + \frac{1}{2} a t^2 \]
Here, \( h \) is the final position (which will be 0 when it hits the ground), \( u \) is the initial velocity, \( a \) is the acceleration (which will be \(-g\) since it acts downwards), and \( t \) is the time.

3. **Set up the equation**:
Since the object is moving upwards initially, we can set the equation as:
\[ 0 = 490 + 400t - \frac{1}{2} \cdot 9.8 t^2 \]
Rearranging gives:
\[ 0 = -\frac{1}{2} \cdot 9.8 t^2 + 400t + 490 \]
Multiplying through by -2 to eliminate the fraction:
\[ 0 = 9.8 t^2 - 800t - 980 \]

4. **Solve the quadratic equation**:
We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 9.8 \), \( b = -800 \), and \( c = -980 \).
\[ b^2 - 4ac = (-800)^2 - 4 \cdot 9.8 \cdot (-980) \]
\[ = 640000 + 38416 = 678416 \]
Now, substituting back into the quadratic formula:
\[ t = \frac{800 \pm \sqrt{678416}}{2 \cdot 9.8} \]
Calculating \( \sqrt{678416} \approx 823.5 \):
\[ t = \frac{800 \pm 823.5}{19.6} \]
This gives two possible solutions for \( t \):
\[ t_1 = \frac{800 + 823.5}{19.6} \quad \text{and} \quad t_2 = \frac{800 - 823.5}{19.6} \]
Calculating \( t_1 \):
\[ t_1 = \frac{1623.5}{19.6} \approx 82.8 \, \text{s} \]
Calculating \( t_2 \) (which will be negative and not physically meaningful):
\[ t_2 = \frac{-23.5}{19.6} \approx -1.2 \, \text{s} \]

5. **Conclusion**:
The time it takes for the second part of the projectile to hit the ground after the breakup is approximately \( 82.8 \, \text{s} \).