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 will follow these steps:

### Step 1: Understand the projectile motion equations
The height \( y \) of a projectile at a horizontal distance \( x \) can be described by the equation:
\[ y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta} \]
where:
- \( y \) is the height of the projectile,
- \( \theta \) is the angle of projection,
- \( g \) is the acceleration due to gravity,
- \( u \) is the initial velocity,
- \( x \) is the horizontal distance from the point of projection.

### Step 2: Substitute the known values
Given:
- \( u = 10 \, \text{m/s} \)
- \( \theta = 45^\circ \)
- \( g = 10 \, \text{m/s}^2 \)
- The height of the hurdles \( y = 2 \, \text{m} \)

Since \( \tan 45^\circ = 1 \) and \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), we can substitute these values into the equation:
\[ 2 = x \cdot 1 - \frac{10 \cdot x^2}{2 \cdot 10^2 \cdot \left(\frac{1}{\sqrt{2}}\right)^2} \]

### Step 3: Simplify the equation
The equation simplifies to:
\[ 2 = x - \frac{10 x^2}{2 \cdot 100 \cdot \frac{1}{2}} = x - \frac{x^2}{10} \]
Rearranging gives:
\[ x - \frac{x^2}{10} - 2 = 0 \]
Multiplying through by 10 to eliminate the fraction:
\[ 10x - x^2 - 20 = 0 \]
Rearranging gives:
\[ -x^2 + 10x - 20 = 0 \]
or
\[ x^2 - 10x + 20 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -10, c = 20 \):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 20}}{2 \cdot 1} \]
\[ x = \frac{10 \pm \sqrt{100 - 80}}{2} \]
\[ x = \frac{10 \pm \sqrt{20}}{2} \]
\[ x = \frac{10 \pm 2\sqrt{5}}{2} \]
\[ x = 5 \pm \sqrt{5} \]

### Step 5: Identify the distances
The two roots are:
- \( x_1 = 5 - \sqrt{5} \) (distance to the first hurdle)
- \( x_2 = 5 + \sqrt{5} \) (distance to the second hurdle)

### Step 6: Calculate the distance \( d \) between the hurdles
The distance \( d \) between the two hurdles is:
\[ d = x_2 - x_1 = (5 + \sqrt{5}) - (5 - \sqrt{5}) = 2\sqrt{5} \]

### Final Answers
- The distance from the point of projection to the first hurdle is \( 5 - \sqrt{5} \, \text{m} \).
- The distance \( d \) between the two hurdles is \( 2\sqrt{5} \, \text{m} \).