Motion in a PlaneNEET Physics · Class 11 · NCERT Chapter 3

A ball is projected with initial speed $20\text{m/s}$ at an angle of $30°$ with the horizontal. Take $g=10{\text{m/s}}^2$. The time of flight is:

A projectile is launched at $45°$ with speed $40\text{m/s}$. Take $g=10{\text{m/s}}^2$. Its maximum height is:

Two stones are projected with the same speed but at angles $30°$ and $60°$ with the horizontal. Their ranges are:

A particle is projected with speed $u$ at an angle $\theta$ with the horizontal. The horizontal component of velocity at the highest point is:

The maximum range of a projectile (on level ground) launched with speed $20\text{m/s}$ is (take $g=10{\text{m/s}}^2$):

At the highest point of a projectile's trajectory, its acceleration is:

A stone is thrown horizontally with speed $10\text{m/s}$ from a tower of height $80\text{m}$. Take $g=10{\text{m/s}}^2$. The horizontal distance covered before hitting the ground is:

From the top of a tower $45\text{m}$ high, a ball is thrown horizontally with speed $20\text{m/s}$. Take $g=10{\text{m/s}}^2$. The speed of the ball just before hitting the ground is:

A particle moves on a circular path of radius $r$ with constant speed $v$. The magnitude of its centripetal acceleration is:

A car travels on a circular track of radius $200\text{m}$ with speed $40\text{m/s}$. Its centripetal acceleration is:

A particle in uniform circular motion has angular velocity $\omega$ and radius $r$. Its centripetal acceleration is:

In uniform circular motion, the work done by the centripetal force is:

Two forces of magnitude $5\text{N}$ and $12\text{N}$ act at a point at right angles to each other. The magnitude of the resultant is:

Two vectors of magnitudes $3$ and $4$ have an angle of $60°$ between them. The magnitude of their resultant is:

If $\vec{A}+\vec{B}=\vec{C}$ and $A=B=C$, the angle between $\vec{A}$ and $\vec{B}$ is:

The dot product $\hat{i}\cdot \hat{j}$ equals:

The cross product $\hat{i}\times \hat{j}$ equals:

If $\vec{A}=2\hat{i}+3\hat{j}$ and $\vec{B}=\hat{i}-\hat{j}$, then $\vec{A}\cdot \vec{B}$ is:

If $\vec{A}$ and $\vec{B}$ have the same magnitude and the angle between them is $\theta$, the magnitude of $\vec{A}+\vec{B}$ equals the magnitude of $\vec{A}-\vec{B}$ when:

The trajectory of a projectile launched at angle $\theta$ with the horizontal is:

The horizontal range and maximum height of a projectile are equal. The angle of projection is:

A ball is thrown at angle $\theta$ with the horizontal. The ratio of the maximum height to the time of flight squared is:

A particle moves in a circle of radius $0.5\text{m}$ with angular velocity $4\text{rad/s}$. Its linear speed and centripetal acceleration are:

A particle moves uniformly on a circular path of radius $r$ with angular speed $\omega$. The time period is:

Two stones are projected horizontally from the top of a cliff with speeds $v_1$ and $v_2$ ($v_2>v_1$). They reach the ground after times:

A projectile is launched at $30°$ with horizontal speed $u$. The ratio of speed at the highest point to the initial speed is:

A vector of magnitude $10$ makes an angle of $60°$ with the x-axis. Its x and y components are:

Two equal vectors of magnitude $A$ have an angle of $120°$ between them. The magnitude of their resultant is:

A projectile lands at the same height from which it was launched. If the initial speed is $u$ and the angle of projection is $\theta$, the time taken to reach the maximum height is:

The time of flight of a projectile is $4\text{s}$. The maximum height reached (take $g=10{\text{m/s}}^2$) is:

The minute hand of a clock has length $5\text{cm}$. The angular speed of its tip is approximately:

A bomb is dropped from an aeroplane flying horizontally at $720\text{km/h}$ at an altitude of $980\text{m}$. Take $g=9.8{\text{m/s}}^2$. The horizontal distance from the release point to where the bomb hits the ground is: