Question 6

The equation of trajectory for a projectile is given by: $y = x \tan \theta - \frac{gx^2}{2v^2 \cos^2 \theta}$

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

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

For option (A) and (B): $\theta = 45^\circ$.

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

$\implies \frac{25g}{v^2} = 4 \implies v^2 = \frac{25g}{4} \implies v = \frac{5\sqrt{g}}{2}$. Thus (A) is correct.

The horizontal distance to maximum height is $x_h = \frac{R}{2} = \frac{v^2 \sin 2\theta}{2g}$.

$x_h = \frac{(25g/4) \cdot \sin 90^\circ}{2g} = \frac{25}{8} = 3.125$ m.

Since $3.125 < 5$, the particle reaches maximum height before reaching $P$. Thus (B) is correct.

For option (C): $\theta = 30^\circ$.

$1 = 5 \tan 30^\circ - \frac{25g}{2v^2 \cos^2 30^\circ} = \frac{5}{\sqrt{3}} - \frac{25g}{2v^2 (3/4)} = \frac{5\sqrt{3}}{3} - \frac{50g}{3v^2}$

$\frac{50g}{3v^2} = \frac{5\sqrt{3}-3}{3} \implies v^2 = \frac{50g}{5\sqrt{3}-3} \approx \frac{50g}{5.66} \approx 8.83g$.

$x_h = \frac{v^2 \sin 60^\circ}{2g} = \frac{8.83g \cdot (\sqrt{3}/2)}{2g} \approx 3.82$ m.

Since $3.82 < 5$, it reaches maximum height before reaching $P$. Thus (C) is incorrect.

For option (D): If $\tan \theta = 1/5$, then $x \tan \theta = 5(1/5) = 1$.

The equation becomes $1 = 1 - \frac{gx^2}{2v^2 \cos^2 \theta}$, which implies $\frac{gx^2}{2v^2 \cos^2 \theta} = 0$. This requires $v \to \infty$. Thus (D) is incorrect.