Motion

Concept:The trajectory of a projectile is given by $y = x \tan \theta - \frac{g x^2}{2 v^2 \cos^2 \theta}$.Explanation:For point P at $(5,1)$, substitute into trajectory equation:$1 = 5 \tan \theta - \frac{25g}{2 v^2 \cos^2 \theta}$.For $\theta = 45^\circ$: $\tan45=1$, $\cos^2 45 = \frac12$.Then $1 = 5 - \frac{25g}{2 v^2 \cdot \frac12} = 5 - \frac{25g}{v^2}$.Solving gives $v^2 = \frac{25g}{4}$, so $v = \frac{5\sqrt{g}}{2}\ \text{ms}^{-1}$. Option A is correct.Find peak position: Range $R = \frac{v^2 \sin 2\theta}{g} = \frac{\frac{25g}{4} \cdot \sin 90^\circ}{g} = 6.25\ \text{m}$.Peak at $x = \frac{R}{2} = 3.125\ \text{m}$. Since $x_P = 5 > 3.125$, particle reaches maximum height before P. Option B is correct.For $\theta = 30^\circ$: Substitute into trajectory equation:$1 = 5 \tan30 - \frac{25g}{2 v^2 \cos^2 30} \Rightarrow 1 = \frac{5}{\sqrt3} - \frac{25g}{2 v^2 \cdot \frac34}$.Solve to get $v^2 \approx 8.83g$.Then $R = \frac{v^2 \sin60^\circ}{g} \approx 7.65\ \text{m}$, peak at $3.825\ \text{m}$. Since $x_P = 5 > 3.825$, peak occurs before P. Option C (peak after P) is false.For $\theta = \tan^{-1}(1/5)$: $\tan\theta = \frac15$, $\cos^2\theta = \frac{1}{1+\tan^2\theta} = \frac{25}{26}$.In trajectory equation: $1 = 5\cdot\frac15 - \frac{25g}{2 v^2 \cdot \frac{25}{26}} = 1 - \frac{13g}{v^2}$.This gives $\frac{13g}{v^2} = 0$, implying $v \to \infty$, not $125\sqrt{g}$. Option D is false.Answer:The correct statements are A and B.