JEE Advanced 2026 — Paper 1 — PHY-II solution

JEE Advanced2026Paper 1PHY-II

A particle is thrown with a speed $v$ from a point $O$ at an angle $θ$ with the horizontal plane such that it passes through the point $P$ at a height of $1$ m and horizontal distance of $5$ m from $O$ , as shown in the figure. If acceleration due to gravity is $g$ ms $^{- 2}$ , then the correct statement(s) is/are :

A
If $θ = 4 5^{\circ}$ , then $v = \frac{5 g}{2} ms^{- 1}$
B
If $θ = 4 5^{\circ}$ , the particle reaches its maximum height before it reaches $P$ .
C
If $θ = 3 0^{\circ}$ , the particle reaches its maximum height after reaching $P$ .
D
If $θ = tan^{- 1} (\frac{1}{5})$ , then $v = 125 g ms^{- 1}$

Solution

Trajectory through P. $y = x tan θ - \frac{g x ^{2}}{2 v ^{2} cos ^{2} θ}$ . With $P = (5, 1)$ : $1 = 5 tan θ - \frac{25 g}{2 v ^{2} cos ^{2} θ}$ .

Option (A): $θ = 4 5^{\circ}$ . $1 = 5 - \frac{25 g}{v ^{2}} \Rightarrow \frac{25 g}{v ^{2}} = 4 \Rightarrow v^{2} = \frac{25 g}{4}$ , so $v = \frac{5 g}{2}$ . Correct.

Option (B): $θ = 4 5^{\circ}$ . Horizontal range $R = \frac{v ^{2} sin 9 0 ^{\circ}}{g} = \frac{25}{4} = 6.25$ m. Apex occurs at $R /2 = 3.125$ m, which is less than 5 m. So the particle has already passed its maximum height before reaching $P$ at $x = 5$ m. Correct.

Option (C): $θ = 3 0^{\circ}$ . Substituting in the trajectory equation gives $v^{2} = \frac{25 g}{4 ( 5 - 3 ) /5}$ , hence $R = \frac{v ^{2} sin 6 0 ^{\circ}}{g}$ works out to about $7.66$ m. Apex at $R /2 \approx 3.83$ m, still less than 5 m. Maximum height is reached before $P$ , not after. Incorrect.

Option (D): $θ = tan^{- 1} (1/5)$ . Then $tan θ = 1/5$ and the line from $O$ to $P$ has the same slope as the launch direction. The required $v$ diverges (the parabola becomes degenerate), so a finite value such as $125 g$ cannot satisfy the equation. Incorrect.

Correct answers: (A) and (B).

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