A particle is projected at an angle $\theta$ with horizontal with an initital speed u. When it makes an angle $\alpha$ with horizontal, its speed v is-

1. $u\cos \theta$

2. $u\cos \theta u\cos \alpha$

3. $\frac{u\sin \theta }{\sin \alpha }$

4. $\frac{u\cos \theta }{\cos \alpha }$

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 $\stackrel{\rightharpoonup }{\mathrm{V}}=k(y\hat{i}+x\hat{j})$; where k is constant. The general equation for 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 body thrown vertically so as to reach its maximum height in t second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is:

(1) $\sqrt{2}t$

(2) $\left(1+\frac{1}{\sqrt{2}}\right)t$

(3) $\frac{3t}{2}$

(4) $\frac{t}{\sqrt{2}}$

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