JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Kinematics

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

Answer: (A), (B)

Solution

Trajectory equation:

y = x tan θ - \frac{1}{2} \frac{gx ^{2}}{v ^{2} c o s ^{2} θ}

. Point P(5,1) lies on trajectory.

(A) For

θ = 4 5^{\circ}

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

gives

v = \frac{5 g}{2}

m/s ✓

(B) Range

R = \frac{v ^{2} s i n 9 0 ^{\circ}}{g} = 6.25

m. Maximum height occurs at

\frac{R}{2} = 3.125

m < 5 m, so particle reaches maximum height before P ✓

(C) For

θ = 3 0^{\circ}

\frac{R}{2} = 3.83

m < 5 m, so particle reaches maximum height before P, not after ✗

(D) For

θ = tan^{- 1} (\frac{1}{5})

: substituting gives

1 = 1 - \frac{g \times 25}{2 v ^{2} \times ( 25/26 )}

, which requires

v \to \infty

✗

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · One or more correct options