A car is moving horizontally along a straight line with a unifrom velocity of `25 m s^-1`. A projectile is to be fired from this car in such a way that it will return to it after it has moved `100 m`. The speed of the projection must be.

Options

A. `10 m s^-1`
B. `20 m s^-1`
C. `15 m s^-1`
D. `25 m s^-1`

Explanation

To solve the problem, we need to determine the speed of the projectile that is fired from a car moving horizontally at a uniform velocity of 25 m/s. The projectile must return to the car after the car has moved 100 m.

### Step-by-step Solution:

1. **Determine the time taken by the car to travel 100 m:**
The car is moving with a uniform velocity of 25 m/s. We can use the formula:
\[ \text{Time} (t) = \frac{\text{Distance}}{\text{Velocity}} = \frac{100 \, \text{m}}{25 \, \text{m/s}} = 4 \, \text{s} \]

2. **Understand the motion of the projectile:**
The projectile is fired from the car and must return to it after the car has traveled 100 m. This means that the projectile will also be in the air for the same amount of time (4 seconds).

3. **Vertical motion of the projectile:**
The vertical motion of the projectile can be analyzed using the following kinematic equation:
\[ \text{Vertical displacement} (y) = u_y t + \frac{1}{2} g t^2 \]
where \(u_y\) is the initial vertical velocity of the projectile, \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)), and \(t\) is the time of flight (4 s).

Since the projectile returns to the same vertical level from which it was fired, the vertical displacement \(y\) will be 0. Therefore, we can set up the equation:
\[ 0 = u_y (4) - \frac{1}{2} g (4^2) \]

4. **Substituting values:**
Substituting \(g = 9.8 \, \text{m/s}^2\):
\[ 0 = u_y (4) - \frac{1}{2} (9.8) (16) \]
\[ 0 = 4u_y - 78.4 \]
\[ 4u_y = 78.4 \]
\[ u_y = \frac{78.4}{4} = 19.6 \, \text{m/s} \]

5. **Finding the speed of projection:**
The speed of projection \(u\) can be found using the Pythagorean theorem, as the projectile has both horizontal and vertical components of velocity:
\[ u = \sqrt{u_x^2 + u_y^2} \]
where \(u_x\) is the horizontal component (which is equal to the speed of the car, 25 m/s) and \(u_y\) is the vertical component (19.6 m/s):
\[ u = \sqrt{(25)^2 + (19.6)^2} \]
\[ u = \sqrt{625 + 384.16} = \sqrt{1009.16} \approx 31.76 \, \text{m/s} \]

### Final Answer:
The speed of the projectile must be approximately **31.76 m/s**.