Kinematics Previous Year Questions (JEE Main 2026)

The horizontal range of a projectile is

$$R = \frac{u^2 \sin 2\theta}{g}$$

With the same $u$ and $g$, the range depends only on $\sin 2\theta$:

$$\frac{R_1}{R_2} = \frac{\sin(2\times 15^\circ)}{\sin(2\times 30^\circ)} = \frac{\sin 30^\circ}{\sin 60^\circ} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$$

So the ratio is $1 : \sqrt{3}$, giving $x = \sqrt{3}$.

Answer: B ($\sqrt{3}$)

The velocity depends on position, so acceleration is found using the chain rule component-wise, $a_i = v_i\,\dfrac{\partial v_i}{\partial x_i}$ (each component of velocity depends only on its own coordinate):

$$a_x = v_x\frac{\partial v_x}{\partial x} = (-x)(-1) = x$$$$a_y = v_y\frac{\partial v_y}{\partial y} = (2y)(2) = 4y$$$$a_z = v_z\frac{\partial v_z}{\partial z} = (-z)(-1) = z$$

At $(1, 2, 4)$:

$$\vec{a} = (1)\hat{i} + (8)\hat{j} + (4)\hat{k}$$$$|\vec{a}| = \sqrt{1^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9\ \text{m/s}^2$$

Answer: B (9)

Work with speeds relative to the ground and take the direction of motion as positive. Car $A$ (ahead) moves at $100$ km/h and car $B$ at $80$ km/h.

The stone is thrown forward from $B$ with speed $v$ relative to $B$, so its ground speed is $(80 + v)$ km/h.

It hits $A$ with a relative speed of $5$ m/s. Convert:

$$5\ \text{m/s} = 5 \times 3.6 = 18\ \text{km/h}$$

The impact speed is the stone’s speed relative to $A$:

$$(80 + v) - 100 = 18$$$$v = 18 + 20 = 38\ \text{km/h}$$

Answer: C (38)

The velocity components are the time derivatives of the coordinates:

$$v_x = \frac{dx}{dt} = 24\ \text{m/s}$$$$v_y = \frac{dy}{dt} = 43.6 - 9.8t$$

At $t = 2$ s:

$$v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24\ \text{m/s}$$

The angle with the horizontal:

$$\tan\theta = \frac{v_y}{v_x} = \frac{24}{24} = 1 \implies \theta = 45^\circ$$

Answer: B (45)

The farthest horizontal distance corresponds to the maximum range, achieved at a launch angle of $45^\circ$:

$$R_{\max} = \frac{u^2}{g}$$

Solving for $u$:

$$u = \sqrt{R_{\max}\,g} = \sqrt{6.4 \times 10} = \sqrt{64} = 8\ \text{m/s}$$

Answer: 8

The ball is dropped from rest, so we need the point where its speed equals the numerical value of $g$, i.e. $v = 10$ m/s.

Using $v^2 = u^2 + 2g\,h_{\text{fall}}$ with $u = 0$:

$$10^2 = 2(10)\,h_{\text{fall}} \implies h_{\text{fall}} = \frac{100}{20} = 5\ \text{m}$$

Height above the ground:

$$h = 18 - 5 = 13\ \text{m}$$

Answer: 13

When released, the stone shares the balloon’s upward velocity, so its initial velocity is $+10$ m/s (upward). Take upward as positive with the origin at the ground.

Stone’s position: $y = 75 + 10t - 5t^2$. It hits the ground when $y = 0$:

$$5t^2 - 10t - 75 = 0 \implies t^2 - 2t - 15 = 0$$$$t = \frac{2 + \sqrt{4 + 60}}{2} = \frac{2 + 8}{2} = 5\ \text{s}$$

Meanwhile the balloon rises at $10$ m/s from $75$ m:

$$h_{\text{balloon}} = 75 + 10 \times 5 = 125\ \text{m}$$

Answer: D (125)

Equal ranges at the same speed occur for complementary launch angles $\theta$ and $90^\circ - \theta$. Their times of flight are

$$T_1 = \frac{2u\sin\theta}{g}, \qquad T_2 = \frac{2u\cos\theta}{g}$$

Multiplying:

$$T_1 T_2 = \frac{4u^2\sin\theta\cos\theta}{g^2} = \frac{2u^2\sin 2\theta}{g^2} = \frac{2R}{g}$$

since $R = \dfrac{u^2\sin 2\theta}{g}$. Solving for $R$:

$$R = \frac{g\,T_1 T_2}{2} = \frac{10 \times 5 \times 10}{2} = 250\ \text{m}$$

Answer: A (250)

The distance covered in the $n$th second is $S_n = u + \dfrac{a}{2}(2n - 1)$. For $n = 5$, this factor is $\dfrac{a}{2}(9) = 4.5a$.

Mass 1 ($u_1 = 5$):

$$5 + 4.5a_1 = 104 \implies a_1 = \frac{99}{4.5} = 22\ \text{m/s}^2$$

Mass 2 ($u_2 = 12$):

$$12 + 4.5a_2 = 129 \implies a_2 = \frac{117}{4.5} = 26\ \text{m/s}^2$$

Velocities after $10$ s using $v = u + at$:

$$v_1 = 5 + 22(10) = 225\ \text{m/s}, \qquad v_2 = 12 + 26(10) = 272\ \text{m/s}$$

Momenta:

$$p_1 = 3.4 \times 225 = 765\ \text{kg·m/s}, \qquad p_2 = 2.5 \times 272 = 680\ \text{kg·m/s}$$$$\frac{p_1}{p_2} = \frac{765}{680} = \frac{9}{8} = \frac{x}{8} \implies x = 9$$

Answer: 9