A cart is moving horizontally along a straight line with constant speed `30 ms^-1.` A particle is to be fired vertically upwards from the moving cart in such a way that it returns to the cart at the same point from where it was projected after the cart has moved `80 m.` At what speed (relative to the cart) must the projectile be fired? (Take `g = 10 ms^-2`)

Options

A. `10 ms^-1`
B. `10 sqrt8 ms^-1`
C. `40/3 ms^-1`
D. None of these

Explanation

To solve the problem, we need to determine the speed at which a particle must be fired vertically upward from a cart moving horizontally at a constant speed of \(30 \, \text{m/s}\), so that it returns to the same point on the cart after the cart has moved \(80 \, \text{m}\).

### Step-by-Step Solution:

1. **Calculate the time taken by the cart to cover 80 m**:
The time taken by the cart can be calculated using the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
Here, the distance is \(80 \, \text{m}\) and the speed of the cart is \(30 \, \text{m/s}\).
\[ \text{Time} = \frac{80 \, \text{m}}{30 \, \text{m/s}} = \frac{8}{3} \, \text{s} \]

2. **Determine the time for the projectile's motion**:
The projectile must return to the cart after the cart has moved \(80 \, \text{m}\). Therefore, the total time of flight for the projectile is also \(\frac{8}{3} \, \text{s}\).

3. **Split the time of flight into ascent and descent**:
Since the time taken to go up is equal to the time taken to come down, we can divide the total time by 2:
\[ \text{Time to go up} = \text{Time to come down} = \frac{4}{3} \, \text{s} \]

4. **Use the kinematic equation to find the initial velocity**:
We can use the kinematic equation:
\[ v = u + at \]
where:
- \(v\) is the final velocity (which is \(0 \, \text{m/s}\) at the peak of the projectile's motion),
- \(u\) is the initial velocity (which we need to find),
- \(a\) is the acceleration (which is \(-g = -10 \, \text{m/s}^2\)),
- \(t\) is the time taken to reach the peak (\(\frac{4}{3} \, \text{s}\)).

Rearranging the equation to solve for \(u\):
\[ 0 = u - 10 \times \frac{4}{3} \]
\[ u = 10 \times \frac{4}{3} = \frac{40}{3} \, \text{m/s} \]

5. **Conclusion**:
The initial speed at which the projectile must be fired relative to the cart is \(\frac{40}{3} \, \text{m/s}\).

### Final Answer:
The speed at which the projectile must be fired relative to the cart is \(\frac{40}{3} \, \text{m/s}\).