A body is projected with velocity $20\sqrt{3}$ m/s with an angle of projection 60$°$ with horizontal. Calculate velocity on that point where body makes an angle 30$°$ with the horizontal.

1. 20 m/s

2. $\frac{20}{\sqrt{3}} m/s$

3. $10\sqrt{3} m/s$

4. 10 m/s

A particle is moving with veocity $\overrightarrow{\mathrm{v}}=\mathrm{k}(\mathrm{y}\hat{\mathrm{i}}+\mathrm{x}\hat{\mathrm{j}})$; where k is constant. The general equation for the path is:

1. $\mathrm{y}={\mathrm{x}}^2+\mathrm{constant}$

2. ${\mathrm{y}}^2={\mathrm{x}}^2+\mathrm{constant}$

3. $\mathrm{y}=\mathrm{x}+\mathrm{constant}$

4. xy=constant

A particle is projected with a velocity u making an angle $\mathrm{\theta }$ with the horizontal. At any instant, its velocity v is at right angles to its initial velocity u; then v is:

1.  u cos $\mathrm{\theta }$

2.  u tan $\mathrm{\theta }$

3.  u cot $\mathrm{\theta }$

4.  u sec $\mathrm{\theta }$

A projectile is given an initial velocity of $\hat{i}+2\hat{j}$. The cartesian equation of its path is (g = 10 ${\mathrm{ms}}^{-2}$) 

1. $\mathrm{y}=2\mathrm{x}-5{\mathrm{x}}^2$

2. $\mathrm{y}=\mathrm{x}-5{\mathrm{x}}^2$

3. $4\mathrm{y}=2\mathrm{x}-5{\mathrm{x}}^2$

4. $\mathrm{y}=2\mathrm{x}-25{\mathrm{x}}^2$

A ship A is moving westwards with a speed of 10 km ${\mathrm{h}}^{-1}$ and a ship B, 100 km south of A is moving northwards with a speed of 10 km ${\mathrm{h}}^{-1}$. The time after which the distance between them becomes the shortest, is:

1. 5 hr

2. $5\sqrt{2}$ hr

3. $10\sqrt{2}$ hr

4. 0 hr

Time taken by the projectile to reach from A to B is t. Then the distance AB is equal to :

1. $\frac{ut}{\sqrt{3}}$

2. $\frac{\sqrt{3}ut}{2}$

3. $\sqrt{3}ut$

4. 2ut

A particle projected with kinetic energy ${\mathrm{k}}_0$ with an angle of projection $\mathrm{\theta }$. Then the variation of kinetic K with vertical displacement y is

1.  linear

2.  parabolic

3.  hyperbolic

4.  periodic

Three particles moving with constant velocities ${\mathrm{V}}_1, {\mathrm{V}}_2$ and V respectively as given in the figure. After some time all three particles are in the same line, then relation among ${\mathrm{V}}_1, {\mathrm{V}}_2$ and V is

1.  $\mathrm{V} = {\mathrm{V}}_1 + {\mathrm{V}}_2$

2.  $\mathrm{V} = \sqrt{{\mathrm{V}}_1{\mathrm{V}}_2}$

3.  $\mathrm{V} = \frac{{\mathrm{V}}_1{\mathrm{V}}_2}{{\mathrm{V}}_1+{\mathrm{V}}_2}$

4.  $\mathrm{V} = \frac{\sqrt{2}{\mathrm{V}}_1{\mathrm{V}}_2}{{\mathrm{V}}_1+{\mathrm{V}}_2}$

A river is flowing with a speed of 1 km/hr. A swimmer wants to go to point 'C' starting from 'A'. He swims with a speed of 5 km/hr, at an angle $\theta$ w.r.t. the river. If AB=BC=400m. Then-

(1) time taken by the man is 12 min

(2) time taken by the man is 8 min

(3) the value of $\theta$ is 45$°$

(4) the value of $\theta$ is 53$°$

 A body is thrown horizontally with a velocity $\sqrt{2gh}$ from the top of a tower of height h. It strikes the level ground through the foot of the tower at a distance x from the tower. The value of x is:

(1) h

(2) $\frac{h}{2}$

(3) 2h

(4) $\frac{2h}{3}$