Equation of trajectory of a projectile is given by `y = -x^(2) + 10x ` where `x` and `y` are in meters and `x` is along horizontal and `y` is vericall `y` upward and particle is projeted from origin. Then : `(g = 10) m//s^(2)`

Options

A. initial velocity of particle is `sqrt(505) m//s`
B. horizontal range is `10 m`
C. maximum height is `25 m`
D. angle of projection with horizontal is `tan^(-1)(5)`.

Explanation

To solve the problem, we need to analyze the given equation of the trajectory of a projectile, which is given by:

\[ y = -x^2 + 10x \]

### Step 1: Find the horizontal range
The horizontal range is the value of \( x \) when \( y = 0 \).

Set \( y = 0 \):
\[ 0 = -x^2 + 10x \]
Rearranging gives:
\[ x^2 - 10x = 0 \]
Factoring out \( x \):
\[ x(x - 10) = 0 \]
Thus, \( x = 0 \) or \( x = 10 \).

So, the horizontal range \( R \) is:
\[ R = 10 \, \text{meters} \]

### Step 2: Find the maximum height
To find the maximum height, we need to determine the vertex of the parabola represented by the equation. The vertex \( x \) coordinate can be found using the formula:
\[ x = -\frac{b}{2a} \]
where \( a = -1 \) and \( b = 10 \):
\[ x = -\frac{10}{2 \cdot -1} = 5 \]

Now, substitute \( x = 5 \) back into the equation to find \( y \):
\[ y = -5^2 + 10 \cdot 5 = -25 + 50 = 25 \]

Thus, the maximum height \( H \) is:
\[ H = 25 \, \text{meters} \]

### Step 3: Find the initial velocity and angle of projection
We can use the formula for maximum height:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Given \( H = 25 \, \text{meters} \) and \( g = 10 \, \text{m/s}^2 \):
\[ 25 = \frac{u^2 \sin^2 \theta}{2 \cdot 10} \]
This simplifies to:
\[ 25 = \frac{u^2 \sin^2 \theta}{20} \]
Multiplying both sides by 20:
\[ 500 = u^2 \sin^2 \theta \]
Thus:
\[ u \sin \theta = \sqrt{500} = 10\sqrt{5} \]

### Step 4: Find the slope at the origin
The slope of the trajectory at any point can be found by differentiating \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = -2x + 10 \]
At the origin \( (0, 0) \):
\[ \frac{dy}{dx} \bigg|_{(0,0)} = -2(0) + 10 = 10 \]

Thus, the slope \( \tan \theta = 10 \), which gives:
\[ \theta = \tan^{-1}(10) \]

### Step 5: Find \( \sin \theta \)
Using the triangle formed by the slope:
\[ \text{Opposite} = 10, \quad \text{Adjacent} = 1 \]
The hypotenuse \( h \) is:
\[ h = \sqrt{10^2 + 1^2} = \sqrt{101} \]
Thus:
\[ \sin \theta = \frac{10}{\sqrt{101}} \]

### Step 6: Substitute back to find \( u \)
Substituting \( \sin \theta \) into \( u \sin \theta = 10\sqrt{5} \):
\[ u \cdot \frac{10}{\sqrt{101}} = 10\sqrt{5} \]
Solving for \( u \):
\[ u = \frac{10\sqrt{5} \cdot \sqrt{101}}{10} = \sqrt{505} \]

### Summary of Results
1. Horizontal Range \( R = 10 \, \text{meters} \)
2. Maximum Height \( H = 25 \, \text{meters} \)
3. Initial Velocity \( u = \sqrt{505} \, \text{m/s} \)
4. Angle of Projection \( \theta = \tan^{-1}(10) \)