JEE Advanced 2026 · Paper 1 · Physics Q6

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PhysicsMultiple correct+4 / -1Question 22 of 48

A particle is thrown with a speed $v$ from a point $O$ at an angle $\theta$ 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:

Point O on a hatched horizontal ground, with velocity v drawn at angle θ above the horizontal. Point P is 1 m above the ground, at a horizontal distance of 5 m from O.

\theta=45^\circ

, then

v=\dfrac{5\sqrt g}2

^{-1}

\theta=45^\circ

, the particle reaches its maximum height before it reaches

P

\theta=30^\circ

, the particle reaches its maximum height after reaching

P

\theta=\tan^{-1}\left(\dfrac15\right)

, then

v=125\sqrt g

^{-1}

Answer and worked solution

Official answer: A, B

Trajectory: $y=x\tan\theta-\dfrac{gx^2}{2v^2\cos^2\theta}$ , with $P=(5,1)$ .

(A) $\theta=45^\circ$ : $1=5-\dfrac{25g}{v^2}$ , so $v^2=\dfrac{25g}4$ and $v=\dfrac{5\sqrt g}2$ . True.

(B) The top is at $x=\dfrac{v^2\sin2\theta}{2g}=\dfrac{25}8=3.125<5$ , before $P$ . True.

(C) $\theta=30^\circ$ : $1=\dfrac5{\sqrt3}-\dfrac{50g}{3v^2}$ , so $v^2\approx8.83g$ . The top is at $x=\dfrac{v^2\sin60^\circ}{2g}\approx3.8<5$ , again before $P$ . False.

(D) $\tan\theta=\frac15$ gives $5\tan\theta=1$ , so the drop term would have to be zero: no finite speed works. False.

Topics:Projectile motion

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