Maximum Height on an Incline

Q. A particle is projected up the incline such that its component of velocity along the incline is $10\text{m/s}$. Time of flight is $2\text{sec}$ and maximum height above the incline is $5\text{m}$. Then velocity of projection will be

1. $10\text{m/s}$
2. $10\sqrt{2}\text{m/s}$
3. $5\sqrt{5}\text{m/s}$
4. $5\sqrt{10}\text{m/s}$

Q. A particle is projected with a velocity of 20 m/s at an angle of ${30}^{\circ }$ to an inclined plane of inclination ${30}^{\circ }$ to the horizontal. The particle hits the inclined plane at an angle ${30}^{\circ }$, during its journey. The time of flight in seconds is -

1. $\frac{4}{\sqrt{3}}$
2. $\frac{2}{\sqrt{3}}$
3. $\sqrt{3}$
4. $\frac{\sqrt{3}}{2}$

Q.

A projectile is projected from the base of an inclined plane of inclination $\alpha$. The angle of projection of the projectile with the ground is $\theta$. If it strikes the incline plane at its maximum height from the ground, then, (assume the plane of motion contains the line of greatest slope on the incline plane)

1. 

2. We can not define a relationship between and as it will depend one the speed of projection

3. 

4. 

Q. A particle is projected up the incline such that its component of velocity along the incline is $10\text{m/s}$. Time of flight is $2\text{sec}$ and maximum height above the incline is $5\text{m}$. Then velocity of projection will be

1. $10\text{m/s}$
2. $10\sqrt{2}\text{m/s}$
3. $5\sqrt{5}\text{m/s}$
4. $5\sqrt{10}\text{m/s}$

Q. A plane is flying horizontally at $98{ms}^{-1}$ and releases an object which reaches the ground in 10 s. The angle made by it while hitting the ground is (in degrees).

1. ${55}^{}$
2. ${45}^{}$
3. ${60}^{}$
4. ${75}^{}$

Q. A jet of water is projected at an angle $\theta ={45}^o$ with horizontal from point A which is situated at a distance $x=OA$ = (a) $1/2$ m, (b) $2$ m from a vertical wall. If the speed of projection is $v_0=\sqrt{1}0m{s}^{-1}$, find point $P$ of striking of the water jet with the vertical wall.

Q. A ball is projected from ground with a speed of 30 m/s , find $\theta$ so that ball passes through point A. $(g=10m{s}^{-2})$

1. ${\tan }^{-1}\left(1\right)$
2. ${\tan }^{-1}\left(3\right)$
3. ${\tan }^{-1}\left(2\right)$
4. ${\tan }^{-1}\left(4\right)$

Q. A particle is projected up the incline such that its component of velocity along the incline is $10\text{m/s}$. Time of flight is $2\text{sec}$ and maximum height above the incline is $5\text{m}$. Then velocity of projection will be

1. $10\text{m/s}$
2. $10\sqrt{2}\text{m/s}$
3. $5\sqrt{5}\text{m/s}$
4. $5\sqrt{10}\text{m/s}$

Q.

A projectile is projected from the base of an inclined plane of inclination '$\alpha$'. The angle of projection of the projectile with the ground is $\theta$. If it strikes the incline plane at it's maximum height from the ground, then, (assume the plane of motion contains the line of greatest slope on the incline plane)

1. tan $\theta$ = $\frac{tan\alpha }{2}$

2. tan $\alpha$ = $\frac{tan\theta }{2}$

3. tan $\alpha$ = $\frac{tan\theta }{3}$

4. We can not define a relationship between $\theta$ and $\alpha$ as it will depend one the speed of projection

Q. A ball is projected from ground with a speed of 30 m/s , find $\theta$ so that ball passes through point A. $(g=10m{s}^{-2})$

1. ${\tan }^{-1}\left(1\right)$
2. ${\tan }^{-1}\left(3\right)$
3. ${\tan }^{-1}\left(2\right)$
4. ${\tan }^{-1}\left(4\right)$

Q. A tank moves uniformly along x-axis. It fires a shot from origin at an angle of ${30}^0$ with horizontal while moving along positive x-axis and the second shot is also fired similarly except that the tank moves along negative x-axis. If the respective range of the shot are $250m$ and $200m$ along x-axis, the velocity of tank is

1. $3.9m{s}^{-1}$
2. $4.9m{s}^{-1}$
3. $9.4m{s}^{-1}$
4. $5.9m{s}^{-1}$

Q. A ball is thrown with initial energy $100$ $J$ at an angle $\theta$ to the horizontal. If its energy at the top becomes $30$ $J$ then angles of projection-

1. $\theta =30$
2. $\theta =45$
3. $\theta =co{s}^{-1}{\left(\frac{3}{10}\right)}^{1/2}$
4. $\theta =co{s}^{-1}\left(\frac{3}{10}\right)$

Q. A particle is projected up the incline such that its component of velocity along the incline is $10\text{m/s}$. Time of flight is $2\text{sec}$ and maximum height above the incline is $5\text{m}$. Then velocity of projection will be

1. $10\text{m/s}$
2. $10\sqrt{2}\text{m/s}$
3. $5\sqrt{5}\text{m/s}$
4. $5\sqrt{10}\text{m/s}$

Q. A ball is projected from ground with a speed of 30 m/s , find $\theta$ so that ball passes through point A. $(g=10m{s}^{-2})$

1. ${\tan }^{-1}\left(1\right)$
2. ${\tan }^{-1}\left(3\right)$
3. ${\tan }^{-1}\left(2\right)$
4. ${\tan }^{-1}\left(4\right)$

Q. A ball is projected from ground with a speed of 30 m/s , find $\theta$ so that ball passes through point A. $(g=10m{s}^{-2})$

1. ${\tan }^{-1}\left(1\right)$
2. ${\tan }^{-1}\left(3\right)$
3. ${\tan }^{-1}\left(2\right)$
4. ${\tan }^{-1}\left(4\right)$