A particle is thrown with a speed v from a point O at an angle θ...

The equation of trajectory of a projectile is given by:

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

The particle passes through point $P(5, 1)$. Substituting $x = 5$ and $y = 1$:

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

For option (A), substituting $\theta = 45^\circ$:

$1 = 5 \tan 45^\circ - \dfrac{25 g}{2 v^2 \cos^2 45^\circ}$

$1 = 5(1) - \dfrac{25 g}{2 v^2 (1/2)}$

$1 = 5 - \dfrac{25 g}{v^2}$

$\dfrac{25 g}{v^2} = 4 \Rightarrow v^2 = \dfrac{25 g}{4} \Rightarrow v = \dfrac{5\sqrt{g}}{2}$ ms$^{-1}$

Thus, option (A) is correct.

For option (B), the horizontal distance to the maximum height is $x_m = \dfrac{R}{2} = \dfrac{v^2 \sin(2\theta)}{2g}$.

For $\theta = 45^\circ$ and $v^2 = \dfrac{25g}{4}$:

$x_m = \dfrac{\left(\dfrac{25g}{4}\right) \sin 90^\circ}{2g} = \dfrac{25}{8} = 3.125$ m

Since $x_m = 3.125$ m $< 5$ m, the particle reaches its maximum height before it reaches $P$. Thus, option (B) is correct.

For option (C), substituting $\theta = 30^\circ$ into the trajectory equation:

$1 = 5 \tan 30^\circ - \dfrac{25 g}{2 v^2 \cos^2 30^\circ}$

$1 = \dfrac{5}{\sqrt{3}} - \dfrac{25 g}{2 v^2 (3/4)} \Rightarrow \dfrac{50 g}{3 v^2} = \dfrac{5}{\sqrt{3}} - 1$

$v^2 = \dfrac{50\sqrt{3}g}{3(5 - \sqrt{3})}$

The horizontal distance to the maximum height is:

$x_m = \dfrac{v^2 \sin 60^\circ}{2g} = \dfrac{v^2 \sqrt{3}}{4g} = \dfrac{50\sqrt{3}g}{3(5 - \sqrt{3})} \times \dfrac{\sqrt{3}}{4g} = \dfrac{25}{2(5 - \sqrt{3})} \approx 3.82$ m

Since $x_m < 5$ m, the particle reaches its maximum height before reaching $P$. Thus, option (C) is incorrect.

For option (D), substituting $\theta = \tan^{-1}\left(\dfrac{1}{5}\right)$, so $\tan \theta = \dfrac{1}{5}$:

$1 = 5\left(\dfrac{1}{5}\right) - \dfrac{25 g}{2 v^2 \cos^2 \theta} \Rightarrow 1 = 1 - \dfrac{25 g}{2 v^2 \cos^2 \theta}$

$\dfrac{25 g}{2 v^2 \cos^2 \theta} = 0$

This is only possible if $v \to \infty$. Thus, option (D) is incorrect.

Answer: If $\theta = 45^\circ$, then $v = \dfrac{5\sqrt{g}}{2}$ ms$^{-1}$.; If $\theta = 45^\circ$, the particle reaches its maximum height before it reaches $P$.