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. `theta_(0)=sin^(-1)((1)/(sqrt(5)))` and `v_(0)=(5)/(3)ms^(-1)`
B. `theta_(0)=cos^(-1)((2)/(sqrt(5)))` and `v_(0)=(3)/(5)ms^(-1)`
C. `theta_(0)=cos^(-1)((1)/(sqrt(5)))` and `v_(0)=(5)/(3)ms^(-1)`
D. `theta_(0)=sin^(-1)((2)/(sqrt(5)))` and `v_(0)=(3)/(5)ms^(-1)`

Answer

To solve the problem, we need to determine the angle of projection (θ) and the speed of projection (u) from the given trajectory equation \( y = 2x - 9x^2 \).

### Step-by-Step Solution:

1. **Identify the Trajectory Equation**:
The given equation is \( y = 2x - 9x^2 \). This is a quadratic equation in the form of \( y = ax + bx^2 \).

2. **Rearranging the Equation**:
We can rewrite the equation as:
\[ y = 2x(1 - \frac{9}{2}x) \]
This shows that the trajectory is a parabola.

3. **Comparing with the General Trajectory Equation**:
The general equation for the trajectory of a projectile is given by:
\[ y = x \tan \theta - \frac{g}{2u^2} x^2 \]
Here, \( g \) is the acceleration due to gravity, and \( u \) is the initial speed of projection.

4. **Identifying Coefficients**:
By comparing coefficients from the two equations:
- From \( y = 2x - 9x^2 \), we can identify:
- \( \tan \theta = 2 \)
- \( \frac{g}{2u^2} = 9 \)

5. **Finding the Angle of Projection**:
To find the angle \( \theta \):
\[ \tan \theta = 2 \]
Thus,
\[ \theta = \tan^{-1}(2) \]

6. **Finding the Speed of Projection**:
We know \( g = 10 \, \text{m/s}^2 \). Using the equation:
\[ \frac{g}{2u^2} = 9 \implies 10 = 18u^2 \implies u^2 = \frac{10}{18} = \frac{5}{9} \]
Therefore,
\[ u = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

7. **Final Results**:
- The angle of projection \( \theta = \tan^{-1}(2) \)
- The speed of projection \( u = \frac{\sqrt{5}}{3} \, \text{m/s} \)

### Summary:
- Angle of projection: \( \theta = \tan^{-1}(2) \)
- Speed of projection: \( u = \frac{\sqrt{5}}{3} \, \text{m/s} \)