Projectile Motion Passing Through a Specific Point

To determine the correct statement(s), we analyze the equation of the trajectory of the projectile. The trajectory equation in terms of horizontal position $x$ and vertical position $y$ is:

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

Given that the particle passes through point $P(5, 1)$, we substitute $x = 5\text{ m}$ and $y = 1\text{ m}$:

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

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Analysis of Options (A) and (B):

Given $\theta = 45^\circ$, we have $\tan 45^\circ = 1$ and $\cos^2 45^\circ = \frac{1}{2}$.

Substituting these values into equation (1): $1 = 5(1) - \frac{25 g}{2 v^2 \left(\frac{1}{2}\right)}$ $1 = 5 - \frac{25 g}{v^2}$ $\frac{25 g}{v^2} = 4 \implies v^2 = \frac{25 g}{4} \implies v = \frac{5\sqrt{g}}{2}\text{ ms}^{-1}$

Thus, Option (A) is correct.

Next, we calculate the horizontal range $R$ for $\theta = 45^\circ$: $R = \frac{v^2 \sin(2\theta)}{g} = \frac{\left(\frac{25 g}{4}\right) \sin(90^\circ)}{g} = \frac{25}{4} = 6.25\text{ m}$

The maximum height is reached at a horizontal distance: $x_{\text{max}} = \frac{R}{2} = \frac{6.25}{2} = 3.125\text{ m}$

Since $x_{\text{max}} = 3.125\text{ m} < 5\text{ m}$ (the $x$-coordinate of point $P$), the particle reaches its maximum height before it reaches point $P$.

Thus, Option (B) is correct.

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Analysis of Option (C):

Given $\theta = 30^\circ$, we have $\tan 30^\circ = \frac{1}{\sqrt{3}}$ and $\cos^2 30^\circ = \frac{3}{4}$.

Substituting into equation (1): $1 = 5\left(\frac{1}{\sqrt{3}}\right) - \frac{25 g}{2 v^2 \left(\frac{3}{4}\right)}$ $1 = \frac{5}{\sqrt{3}} - \frac{50 g}{3 v^2} \implies \frac{50 g}{3 v^2} = \frac{5 - \sqrt{3}}{\sqrt{3}}$ $v^2 = \frac{50 \sqrt{3} g}{3(5 - \sqrt{3})}$

The horizontal range $R$ is: $R = \frac{v^2 \sin(60^\circ)}{g} = \frac{50 \sqrt{3}}{3(5 - \sqrt{3})} \cdot \frac{\sqrt{3}}{2} = \frac{25}{5 - \sqrt{3}} \approx 7.65\text{ m}$

The horizontal distance to reach maximum height is: $x_{\text{max}} = \frac{R}{2} \approx \frac{7.65}{2} = 3.825\text{ m}$

Since $x_{\text{max}} \approx 3.825\text{ m} < 5\text{ m}$, the particle reaches its maximum height before reaching point $P$.

Thus, Option (C) is incorrect.

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Analysis of Option (D):

Given $\theta = \tan^{-1}\left(\frac{1}{5}\right)$, we have $\tan\theta = \frac{1}{5}$.

Substituting into equation (1): $1 = 5\left(\frac{1}{5}\right) - \frac{25 g}{2 v^2 \cos^2\theta}$ $1 = 1 - \frac{25 g}{2 v^2 \cos^2\theta} \implies \frac{25 g}{2 v^2 \cos^2\theta} = 0$

This equation has no solution for any finite launch speed $v$. Geometrically, the line connecting $O(0,0)$ and $P(5,1)$ forms an angle of $\tan^{-1}(1/5)$ with the horizontal. Due to downward gravitational acceleration, a projectile launched at this angle will always curve below this line and can never pass through point $P$ for any finite speed.

Thus, Option (D) is incorrect.

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Conclusion:

The correct options are A and B.