Let particle be projected from the origin $O(0,0)$ with an initial velocity $v$ at an angle $\theta$ with the horizontal.
The horizontal component of velocity is constant is $v_x = v \cos \theta$
The vertical component of velocity changes due to gravity is $v_y = v \sin \theta - gt$
The position coordinates at any time $t$ are given by :
Horizontal displacement :
$x = (v \cos \theta)t \Rightarrow t = \frac{x}{v \cos \theta}$
Vertical displacement :
$y = (v \sin \theta)t - \frac{1}{2}gt^2$
Substituting the expression for $t$ from the horizontal motion into the vertical displacement equation :
$y = (v \sin \theta)\left(\frac{x}{v \cos \theta}\right) - \frac{1}{2}g\left(\frac{x}{v \cos \theta}\right)^2$
$y = x \tan \theta - \frac{gx^2}{2v^2} \cos^2 \theta$
As the particle must pass through point P with coordinates $\mathrm{x}=5 \mathrm{~m}$ and $\mathrm{y}=1 \mathrm{~m}$.
Substituting these coordinates into our trajectory equation :
$ 1=5 \tan \theta-\frac{g(5)^2}{2 v^2 \cos ^2 \theta} $
$\Rightarrow $ $ 1=5 \tan \theta-\frac{25 g}{2 v^2 \cos ^2 \theta} $
When $\theta=45^{\circ}$
$ 1=5 \tan 45^{\circ}-\frac{25 \mathrm{~g}}{2 \mathrm{v}^2 \cos ^2 45^{\circ}} $
$\Rightarrow $ $1=5 \times 1-\frac{25 \mathrm{~g}}{2 \mathrm{v}^2\left(\frac{1}{\sqrt{2}}\right)^2}$
$\Rightarrow $ $\frac{25 \mathrm{~g}}{\mathrm{v}^2}=4$
$\Rightarrow $ $v^2=\frac{25 g}{4}$
$\Rightarrow $ $\mathrm{v}=\frac{5 \sqrt{\mathrm{~g}}}{2} \mathrm{~ms}^{-1}$
Thus, option (A) is correct.
For a projectile, the horizontal distance from the point of projection to the peak of its trajectory is exactly half of its total horizontal range (R).
The formula for the horizontal range is $\mathrm{R}=\frac{\mathrm{v}^2 \sin (2 \theta)}{\mathrm{g}}$
Substituting $\mathrm{v}^2=\frac{25 \mathrm{~g}}{4}$ and $\theta=45^{\circ}$ :
$ \mathrm{R}=\frac{\left(\frac{25 \mathrm{~g}}{4}\right) \sin \left(90^{\circ}\right)}{\mathrm{g}}=\frac{25}{4}=6.25 \mathrm{~m} $
The horizontal coordinate of the maximum height is:
$ \mathrm{x}_{\mathrm{H}_{\max }}=\frac{\mathrm{R}}{2}=\frac{6.25}{2}=3.125 \mathrm{~m} $
Since the horizontal coordinate of point P is $\mathrm{x}_{\mathrm{P}}=5 \mathrm{~m}$, and $5 \mathrm{~m}>3.125 \mathrm{~m}\left(\mathrm{x}_{\mathrm{P}}>\mathrm{x}_{\mathrm{H}_{\max }}\right)$, the particle reaches its peak position horizontally at 3.125 m before continuing downward to pass through point P at $\mathrm{x}=5$ m. Thus, option (B) is correct.
We need to find the horizontal coordinate of the peak position $\left(\mathrm{x}_{\mathrm{H}_{\max }}\right)$ when $\theta=30^{\circ}$ to determine whether the peak occurs before or after $P$.
For $\theta=30^{\circ}$ using equation of trajectory,
$ \mathrm{y}=\mathrm{x} \tan \theta-\frac{\mathrm{gx}^2}{2 \mathrm{v}^2 \cos ^2 \theta} $
$\Rightarrow $ $1=5 \times \tan 30^{\circ}-\frac{\mathrm{g} \times 5^2}{2 \mathrm{v}^2 \cos ^2 30^{\circ}}$
$\Rightarrow $ $1=5\left(\frac{1}{\sqrt{3}}\right)-\frac{25 \mathrm{~g}}{2 \mathrm{v}^2\left(\frac{\sqrt{3}}{2}\right)^2}$
$\Rightarrow $ $1=\frac{5}{\sqrt{3}}-\frac{50 g}{3 v^2}$
$\Rightarrow $ $\frac{50 \mathrm{~g}}{3 \mathrm{v}^2}=\frac{5}{\sqrt{3}}-1 \approx 51.732-1 \approx 2.887-1=1.887$
$\Rightarrow $ $ \mathrm{v}^2=\frac{50 \mathrm{~g}}{3 \times 1.887} \approx 8.83 \mathrm{~g} $
The total horizontal range R :
$ \mathrm{R}=\frac{\mathrm{v}^2 \sin (2 \theta)}{\mathrm{g}}=\frac{(8.83 \mathrm{~g}) \sin \left(60^{\circ}\right)}{\mathrm{g}}=8.83 \times \frac{\sqrt{3}}{2} \approx 8.83 \times 0.866 \approx 7.65 \mathrm{~m} $
The horizontal coordinate of the peak is:
$ \mathrm{x}_{\mathrm{H}_{\max }}=\frac{\mathrm{R}}{2} \approx 7.652=3.825 \mathrm{~m} $
As $\mathrm{x}_{\mathrm{P}}=5 \mathrm{~m}, \mathrm{x}_{\mathrm{H}_{\max }}<5 \mathrm{~m}$ which means the particle reaches its maximum height at 3.825 m before it reaches point $\mathrm{P}(\mathrm{x}=5 \mathrm{~m})$.
In option (C) it says its maximum height after reaching P , so option (C) is incorrect.
When $\theta=\tan ^{-1}\left(\frac{1}{5}\right) \Rightarrow \tan \theta=\frac{1}{5}$ using the trigonometric identity $\cos ^2 \theta=\frac{1}{1+\tan ^2 \theta}$ :
$ \cos ^2 \theta=\frac{1}{1+\left(\frac{1}{5}\right)^2}=\frac{1}{1+\frac{1}{25}}=\frac{25}{26} $
Substituting $\tan \theta=\frac{1}{5}$ and $\cos ^2 \theta=\frac{25}{26}$ into trajectory equation:
$ 1=5\left(\frac{1}{5}\right)-\frac{25 \mathrm{~g}}{2 \mathrm{v}^2(25 / 26)} $
$\Rightarrow $ $1=1-\frac{26 g}{2 v^2}$
$\Rightarrow $ $1=1-\frac{13 g}{v^2}$
$\Rightarrow $ $ 0=-\frac{13 g}{v^2} \Rightarrow \frac{13 g}{v^2}=0 \Rightarrow v \rightarrow \infty $
Thus, option (D) is incorrect.
Therefore, the correct options are (A) and (B).
