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 step by step, let's break it down:

### Step 1: Understand the motion of the cart and the projectile
The cart is moving horizontally with a constant speed of \(30 \, \text{m/s}\). A particle is fired vertically upward from the cart and must return to the same point on the cart after the cart has moved \(80 \, \text{m}\).

### Step 2: Calculate the time taken for the cart to move 80 m
Using the formula for distance:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
We can rearrange this to find the time:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{80 \, \text{m}}{30 \, \text{m/s}} = \frac{8}{3} \, \text{s} \]

### Step 3: Analyze the vertical motion of the projectile
The projectile is fired vertically upwards with an initial speed \(u\) (relative to the cart). The time of flight for the projectile to go up and come back down is given by:
\[ T = \frac{2u}{g} \]
where \(g\) is the acceleration due to gravity, which is \(10 \, \text{m/s}^2\).

### Step 4: Set the time of flight equal to the time taken by the cart
Since the projectile must return to the cart after the same time the cart takes to move \(80 \, \text{m}\), we set:
\[ \frac{2u}{g} = \frac{8}{3} \]
Substituting \(g = 10 \, \text{m/s}^2\):
\[ \frac{2u}{10} = \frac{8}{3} \]

### Step 5: Solve for \(u\)
Multiplying both sides by \(10\):
\[ 2u = \frac{80}{3} \]
Now, divide by \(2\):
\[ u = \frac{80}{6} = \frac{40}{3} \, \text{m/s} \]

### Step 6: Conclusion
The speed at which the projectile must be fired relative to the cart is:
\[ u = \frac{40}{3} \, \text{m/s} \approx 13.33 \, \text{m/s} \]

### Final Answer
The projectile must be fired upwards at a speed of approximately \(13.33 \, \text{m/s}\) relative to the cart.

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