A particle is thrown with a speed $v$ from a point $O$ at an angle $\theta$ with the horizontal plane such that it passes through the point $P$ at a height of $1\,\text{m}$ and horizontal distance of $5\,\text{m}$ from $O$, as shown in the figure. If acceleration due to gravity is $g\,\text{ms}^{-2}$, then the correct statement(s) is/are:

Step 1: Write the equation of trajectory.

The trajectory of a projectile launched from the origin is given by the equation:

$$y = x\tan\theta - \frac{g x^2}{2v^2\cos^2\theta}$$

Step 2: Substitute the coordinates of point $P$.

Since the particle passes through $P(5, 1)$, we substitute $x = 5$ and $y = 1$.

$$1 = 5\tan\theta - \frac{25g}{2v^2\cos^2\theta}$$

Rearranging this equation yields the expression for $v^2$.

$$v^2 = \frac{25g}{2\cos\theta(5\sin\theta - \cos\theta)}$$

Step 3: Evaluate Statement (1) for $\theta = 45^\circ$.

Substitute $\theta = 45^\circ$ into the expression for $v^2$.

$$v^2 = \frac{25g}{2\cos 45^\circ(5\sin 45^\circ - \cos 45^\circ)}$$ $$v^2 = \frac{25g}{4}$$ $$v = \frac{5\sqrt{g}}{2}$$

Statement (1) is correct.

Step 4: Evaluate Statement (2) for $\theta = 45^\circ$.

The horizontal coordinate of the peak of the trajectory is given by:

$$x_{\text{peak}} = \frac{v^2\sin\theta\cos\theta}{g}$$

Substitute $\theta = 45^\circ$ and $v^2 = \frac{25g}{4}$.

$$x_{\text{peak}} = \frac{25}{8}$$ $$x_{\text{peak}} = 3.125\,\text{m}$$

Since $x_{\text{peak}} < 5\,\text{m}$, the particle reaches its maximum height before reaching $P$, so Statement (2) is correct.

Step 5: Evaluate Statement (3) for $\theta = 30^\circ$.

Substitute $\theta = 30^\circ$ into the expression for $v^2$.

$$v^2 = \frac{50g}{5\sqrt{3}-3}$$

Calculate the horizontal coordinate of the peak.

$$x_{\text{peak}} = \frac{25}{2(5-\sqrt{3})}$$ $$x_{\text{peak}} \approx 3.83\,\text{m}$$

Since $x_{\text{peak}} < 5\,\text{m}$, the particle reaches its maximum height before reaching $P$, so Statement (3) is incorrect.

Step 6: Evaluate Statement (4) for $\theta = \tan^{-1}(1/5)$.

For $\theta = \tan^{-1}(1/5)$, the denominator of the expression for $v^2$ becomes zero.

$$5\sin\theta - \cos\theta = 0$$ $$v^2 \to \infty$$

Statement (4) is incorrect.

Answer: Correct statements are (1) and (2).