The front wind screen of a car is inclined at an `60^@` with the vertical. Hailstones fall vertically downwards with a speed of `5 sqrt(3) ms^-1`. Find the speed of the car so that hailstones are bounced back by the screen in vertically upward direction with respect to car. Assume elastic collision of hailstones with wind screen.

Explanation

To solve the problem, we need to find the speed of the car such that hailstones, which are falling vertically downward, bounce off the inclined windscreen of the car and move vertically upward with respect to the car.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The windscreen is inclined at an angle of \(60^\circ\) with the vertical.
- Hailstones are falling vertically downwards with a speed of \(5\sqrt{3} \, \text{m/s}\).
- We want the hailstones to bounce back vertically upwards with respect to the car after colliding with the windscreen.

2. **Setting Up the Angles**:
- Since the windscreen is inclined at \(60^\circ\) with the vertical, it makes an angle of \(30^\circ\) with the horizontal (since \(90^\circ - 60^\circ = 30^\circ\)).
- After bouncing off the windscreen, we want the hailstones to move vertically upwards, which means they will make an angle of \(30^\circ\) with the horizontal.

3. **Using the Concept of Elastic Collision**:
- In an elastic collision, the angle of incidence is equal to the angle of reflection. Therefore, if the hailstones hit the windscreen at an angle of \(30^\circ\), they will bounce off at the same angle of \(30^\circ\).

4. **Velocity Components**:
- Let \(v_c\) be the speed of the car.
- The velocity of the hailstones \(v_h = 5\sqrt{3} \, \text{m/s}\) is directed vertically downward.
- The velocity of the hailstones with respect to the car can be expressed as:
\[ v_{hc} = v_h - v_c \]
- We need to find \(v_c\) such that \(v_{hc}\) makes an angle of \(30^\circ\) with the horizontal.

5. **Resolving the Hailstone's Velocity**:
- The vertical component of the hailstone's velocity with respect to the car:
\[ v_{hc_y} = v_h - v_c \]
- The horizontal component of the hailstone's velocity with respect to the car:
\[ v_{hc_x} = v_{hc} \cdot \tan(30^\circ) = v_{hc} \cdot \frac{1}{\sqrt{3}} \]

6. **Using the Right Triangle**:
- From the triangle formed by the velocities, we have:
\[ \tan(30^\circ) = \frac{v_{hc_y}}{v_{hc_x}} \]
- Substituting the known values:
\[ \frac{1}{\sqrt{3}} = \frac{(5\sqrt{3} - v_c)}{v_c} \]

7. **Solving for \(v_c\)**:
- Cross-multiplying gives:
\[ v_c = \sqrt{3}(5\sqrt{3} - v_c) \]
- Rearranging:
\[ v_c + \sqrt{3}v_c = 15 \]
\[ v_c(1 + \sqrt{3}) = 15 \]
\[ v_c = \frac{15}{1 + \sqrt{3}} \]
- Rationalizing the denominator:
\[ v_c = \frac{15(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{15(1 - \sqrt{3})}{1 - 3} = \frac{15(1 - \sqrt{3})}{-2} \]
\[ v_c = -\frac{15(1 - \sqrt{3})}{2} \]
- Since speed cannot be negative, we take the absolute value:
\[ v_c = \frac{15(\sqrt{3} - 1)}{2} \]

8. **Final Calculation**:
- Approximating \(\sqrt{3} \approx 1.732\):
\[ v_c \approx \frac{15(1.732 - 1)}{2} \approx \frac{15(0.732)}{2} \approx \frac{10.98}{2} \approx 5.49 \, \text{m/s} \]

### Conclusion:
The speed of the car required for the hailstones to bounce back vertically upward with respect to the car is approximately \(5.49 \, \text{m/s}\).