Through-the-Ring Launcher — a projectile through a point — JEE Advanced 2026 Paper 1 Q.6

01The question

Q.6 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 = 45°, then v = 5√g/2 ms⁻¹.
(B)If θ = 45°, the particle reaches its maximum height before it reaches P.
(C)If θ = 30°, the particle reaches its maximum height after reaching P.
(D)If θ = tan⁻¹(1/5), then v = 125√g ms⁻¹.

02The experiment

Answer— Press START: the launcher throws at each angle in the options and the throws are measured.

Ready.

Angle θ

Any angle45°

Speed

03Detailed solution

Answer: —

The path. With the origin at O:
y = x tan θ − g x²/(2v² cos²θ).
Passing through P (5 m, 1 m) fixes the speed for each angle:
v² = g·25 / [2 cos²θ (5 tan θ − 1)].
(A) is true. At θ = 45°: v² = 25g/[2·½·(5 − 1)] = 25g/4, so v = 5√g/2 m/s.
(B) is true. The highest point is at x = v² sin 2θ/(2g) = (25/4)/2 = 3.125 m, which is before P at 5 m. The ball is already coming down when it passes P.
(C) is false. At θ = 30°: v² = 25g/[2·¾·(5/√3 − 1)] ≈ 8.83g. The highest point is at x = 8.83·(√3/2)/2 ≈ 3.82 m, still before P, not after it.
(D) is false. tan θ = 1/5 is the slope of the straight line OP itself, so 5 tan θ − 1 = 0 and v → ∞. Gravity bends every real path below the line of sight. With v = 125√g the ball passes 0.83 mm below P, and no finite speed reaches it.

Angle	Speed needed to pass P	Highest point at	Relative to P
45°	2.500 √g m/s	3.125 m	before P
30°	2.972 √g m/s	3.825 m	before P
tan⁻¹(1/5) ≈ 11.3°	∞ (no finite speed)	—	unreachable

Understand the concept

For a given launch angle, only one speed makes the parabola pass through a given point, and it comes from the trajectory equation. The highest point is where the vertical velocity is zero, at x = v² sin 2θ/2g. Comparing that with 5 m tells you whether the ball is still rising at P. Aiming straight at a point (tan θ = height/distance) never works: gravity pulls the ball below the line of sight at any finite speed.

Key takeaways

Trajectory: y = x tan θ − g x²/(2v² cos²θ). Put in the point to find the speed.
The highest point is at x = v² sin 2θ/2g, half the range on level ground.
The highest point here is at x = 12.5 tan θ/(5 tan θ − 1). It lies beyond P only for shallow throws, tan⁻¹(1/5) < θ < tan⁻¹(2/5) (about 11.3° to 21.8°). Try 20° with the slider.
A throw aimed along the line of sight to P needs infinite speed.

Verification: an independent derivation, a sympy route, a numerical integration of the motion (no trajectory formula), and the page's own engine, run by tests/verify_p1phyq06.py. The official key (A, B) and the MathonGo solution (A, B) agree. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded. The scene is drawn to scale in metres and air resistance is neglected. The 0.83 mm miss in (D) is drawn exaggerated.

How to use this simulation

Press ▶ START EXPERIMENT. The launcher throws at 45°, then 30°, then aims along tan⁻¹(1/5), each at the speed needed to pass P.
Watch the flag at each throw's highest point: is it before or after the ring P?
For tan⁻¹(1/5), watch faster and faster throws fall below P. The last is option (D)'s 125√g.
Tap an Angle θ button or drag Any angle to throw one angle yourself and read v and the highest point.
Replay runs it again. Reset clears it. Classroom fills the screen for a projector.