A particle is projected from origin with velocity $\vec{v} = (3\hat{i} + 4\hat{j})$ m/s. Take $g = 10\hat{j}$ (downward). The trajectory equation $y(x)$ is:
A particle moves with acceleration $a = -kv^2$. If initial velocity is $u$, the velocity after traveling distance $x$ is:
A projectile is fired with velocity $u$ at angle $\theta$ on an inclined plane of inclination $\alpha$ (up the plane). The range along the incline is:
A man can swim at $5$ km/h in still water. He wants to cross a river $1$ km wide flowing at $3$ km/h, in the shortest possible time. He should swim:
A ball is dropped from a tower. During the last second of fall, it covers $7/16$ of the total height. The height of the tower ($g = 10 \text{ m/s}^2$):
A particle starts from rest with constant angular acceleration $2 \text{ rad/s}^2$ on a circle of radius $5$ m. The magnitude of net acceleration after $1$ s is:
From the top of a building of height $40$ m, an object is thrown with velocity $20$ m/s at $60^\circ$ above horizontal. Time of flight ($g = 10 \text{ m/s}^2$):
Two particles A and B are projected from same point with velocities $v_1$ and $v_2$ at angles $\theta_1$ and $\theta_2$ such that they have same horizontal range. If $\theta_1 = 30^\circ$ and $\theta_2 = 60^\circ$, then $v_1/v_2$:
A wheel rotates with $\omega = 5 + 4t$ rad/s. The angle through which the wheel rotates in $2$ s starting from $t = 0$:
A stone tied to a string is whirled in a vertical circle of radius $r$. Minimum speed at the top for the string to remain taut:
A particle moves on a straight line with $v = 2x$ (in SI). If $x = 1$ at $t = 0$, find $x$ at $t = \ln 4$ seconds.
A projectile is launched at $60^\circ$ with $u = 40$ m/s. The horizontal distance covered when the velocity makes $45^\circ$ with horizontal (m, $g = 10$):
A boat is sailing with $4$ m/s relative to water in the east direction. River flows south at $3$ m/s. A boy walks across the boat at $\sqrt{6}$ m/s perpendicular to its length (boat aligned east-west). His speed relative to ground (m/s):
Assertion (A): In uniform circular motion, work done by centripetal force is zero in one revolution. Reason (R): Centripetal force is always perpendicular to velocity.
Assertion (A): The horizontal range of projectile at angle $\theta$ from horizontal is same as that at $(90 - \theta)$. Reason (R): $\sin 2\theta = \sin(180 - 2\theta) = \sin(2(90-\theta))$.
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