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. `(20)/(3)ms^(-1)`
C. `(40)/(3)ms^(-1)`
D. `(80)/(3)ms^(-1)`

Explanation

To solve the problem, we need to determine the speed at which a particle must be fired vertically upwards from a moving cart so that it returns to the cart after the cart has moved a distance of 80 meters.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Speed of the cart, \( v_c = 30 \, \text{m/s} \)
- Distance moved by the cart, \( d = 80 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Time Taken for the Cart to Move 80 m:**
- The time \( t \) taken for the cart to move a distance \( d \) can be calculated using the formula:
\[ t = \frac{d}{v_c} \]
- Substituting the known values:
\[ t = \frac{80 \, \text{m}}{30 \, \text{m/s}} = \frac{8}{3} \, \text{s} \]

3. **Understanding the Vertical Motion of the Particle:**
- The particle is fired vertically upwards and must return to the same point on the cart after time \( t \).
- Since the displacement in the vertical direction is zero (it returns to the same height), we can use the kinematic equation:
\[ s = ut + \frac{1}{2}(-g)t^2 \]
- Here, \( s = 0 \) (displacement), \( u \) is the initial velocity of the particle relative to the cart, and \( g \) is the acceleration due to gravity (acting downwards).

4. **Set Up the Equation:**
- Substituting \( s = 0 \) into the equation gives:
\[ 0 = ut - \frac{1}{2}gt^2 \]
- Rearranging this gives:
\[ ut = \frac{1}{2}gt^2 \]
- Therefore, we can express \( u \) as:
\[ u = \frac{1}{2}g \cdot t \]

5. **Substituting the Values:**
- Now substituting \( g = 10 \, \text{m/s}^2 \) and \( t = \frac{8}{3} \, \text{s} \):
\[ u = \frac{1}{2} \cdot 10 \cdot \frac{8}{3} \]
- Simplifying this:
\[ u = \frac{10 \cdot 8}{2 \cdot 3} = \frac{80}{6} = \frac{40}{3} \, \text{m/s} \]

6. **Conclusion:**
- The speed at which the particle must be fired relative to the cart is:
\[ u = \frac{40}{3} \, \text{m/s} \]

### Final Answer:
The required speed relative to the cart is \( \frac{40}{3} \, \text{m/s} \).

---