Equation of trajector of ground to ground projectile is `y=2x-9x^(2)`. Then the angle of projection with horizontal and speed of projection is : `(g=10m//s^(2))`

Options

A. `tan^(-1)(2),(5)/(2)m//s`
B. `tan^(-1)(3),(5)/(3)m//s`
C. `tan^(-1)(2),(2)/(3),(2)/(3)m//s`
D. `tan^(-1)(3),(2)/(3)m//s`

Answer

To solve the problem of finding the angle of projection and speed of projection for the given trajectory equation \( y = 2x - 9x^2 \), we can follow these steps:

### Step 1: Identify the form of the trajectory equation
The general form of the trajectory equation for a projectile is given by:
\[ y = x \tan \theta - \frac{g}{2u^2 \cos^2 \theta} x^2 \]
where \( \theta \) is the angle of projection, \( u \) is the initial speed, and \( g \) is the acceleration due to gravity.

### Step 2: Compare the given equation with the general form
From the question, we have:
\[ y = 2x - 9x^2 \]
We can rewrite this in the form:
\[ y = 2x - \left(9\right)x^2 \]
Comparing this with the general form, we can identify:
- \( \tan \theta = 2 \)
- \( \frac{g}{2u^2 \cos^2 \theta} = 9 \)

### Step 3: Calculate the angle of projection
To find the angle \( \theta \), we can use the relation:
\[ \tan \theta = 2 \]
Thus,
\[ \theta = \tan^{-1}(2) \]

### Step 4: Calculate \( \cos \theta \) and \( \sin \theta \)
Using the triangle formed by the tangent:
- Opposite side = 2
- Adjacent side = 1
- Hypotenuse = \( \sqrt{2^2 + 1^2} = \sqrt{5} \)

Now, we can find:
\[ \sin \theta = \frac{2}{\sqrt{5}}, \quad \cos \theta = \frac{1}{\sqrt{5}} \]

### Step 5: Substitute into the range equation
From the comparison, we have:
\[ \frac{g}{2u^2 \cos^2 \theta} = 9 \]
Substituting \( g = 10 \, \text{m/s}^2 \) and \( \cos^2 \theta = \left(\frac{1}{\sqrt{5}}\right)^2 = \frac{1}{5} \):
\[ \frac{10}{2u^2 \cdot \frac{1}{5}} = 9 \]
This simplifies to:
\[ \frac{10 \cdot 5}{2u^2} = 9 \]
\[ \frac{50}{2u^2} = 9 \]
\[ \frac{25}{u^2} = 9 \]
\[ u^2 = \frac{25}{9} \]
\[ u = \frac{5}{3} \, \text{m/s} \]

### Final Answers
- The angle of projection \( \theta = \tan^{-1}(2) \)
- The speed of projection \( u = \frac{5}{3} \, \text{m/s} \)