A projectile takes off with an initial velocity of `10 m//s` at an angle of elevation of `45^@`. It is just able to clear two hurdles of height 2 m each, separated from each other by a distance d. Calculate d. At what distance from the point of projection is the first hurdle placed? Take `g = 10m//s^2`.

Explanation

To solve the problem, we need to find the distance \( d \) between the two hurdles and the distance from the point of projection to the first hurdle.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial velocity \( u = 10 \, \text{m/s} \)
- Angle of elevation \( \theta = 45^\circ \)
- Height of hurdles \( y = 2 \, \text{m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

2. **Resolve the initial velocity into horizontal and vertical components:**
\[ u_x = u \cos \theta = 10 \cos 45^\circ = 10 \times \frac{1}{\sqrt{2}} = 5\sqrt{2} \, \text{m/s} \]
\[ u_y = u \sin \theta = 10 \sin 45^\circ = 10 \times \frac{1}{\sqrt{2}} = 5\sqrt{2} \, \text{m/s} \]

3. **Use the projectile motion equation to find the horizontal distance \( x \) when the projectile reaches the height of the hurdles:**
The equation for vertical motion is given by:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
Substituting the known values:
\[ 2 = x \tan 45^\circ - \frac{10 x^2}{2 \times 10^2 \cos^2 45^\circ} \]
Since \( \tan 45^\circ = 1 \) and \( \cos^2 45^\circ = \frac{1}{2} \):
\[ 2 = x - \frac{10 x^2}{2 \times 100 \times \frac{1}{2}} \]
Simplifying gives:
\[ 2 = x - \frac{x^2}{10} \]

4. **Rearranging the equation:**
\[ \frac{x^2}{10} + 2 - x = 0 \]
Multiplying through by 10 to eliminate the fraction:
\[ x^2 - 10x + 20 = 0 \]

5. **Using the quadratic formula to solve for \( x \):**
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -10, c = 20 \):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 20}}{2 \cdot 1} \]
\[ x = \frac{10 \pm \sqrt{100 - 80}}{2} \]
\[ x = \frac{10 \pm \sqrt{20}}{2} \]
\[ x = \frac{10 \pm 2\sqrt{5}}{2} \]
\[ x = 5 \pm \sqrt{5} \]

6. **Finding the distances:**
- The two possible values for \( x \) are \( 5 + \sqrt{5} \) and \( 5 - \sqrt{5} \).
- The distance \( d \) between the two hurdles is:
\[ d = (5 + \sqrt{5}) - (5 - \sqrt{5}) = 2\sqrt{5} \]
- The distance from the point of projection to the first hurdle is:
\[ x_1 = 5 - \sqrt{5} \]

7. **Final calculations:**
- Calculate \( d \):
\[ d \approx 2 \times 2.236 = 4.472 \, \text{m} \approx 4.47 \, \text{m} \]
- Calculate \( x_1 \):
\[ x_1 \approx 5 - 2.236 = 2.764 \, \text{m} \approx 2.75 \, \text{m} \]

### Final Answers:
- The distance \( d \) between the hurdles is approximately \( 4.47 \, \text{m} \).
- The distance from the point of projection to the first hurdle is approximately \( 2.75 \, \text{m} \).