The radius vector of a particle moving on a circle is given by $\overrightarrow{\mathrm{r}} = \mathrm{AcosBt}\hat{\mathrm{i}} + \mathrm{AsinBtj}\stackrel{}{^}$ (A and B are constants). The radius of the circle and speed of the particle, respectively, are

1.  A, AB

2.  $\mathrm{A}, \frac{{\mathrm{A}}^2}{\mathrm{B}}$

3.  B, AB

4.  $\mathrm{B}, \frac{{\mathrm{A}}^2}{\mathrm{B}}$

For a particle projected on a horizontal surface, the ratio of range to maximum height is:

($\mathrm{\theta }$ is the angle of projection with respect to horizontal):

(1) 4cot$\mathrm{\theta }$

(2) 4tan$\mathrm{\theta }$

(3) 0.25cot$\mathrm{\theta }$

(4) 0.25sin$\mathrm{\theta }$

Two projectiles, one fired from the surface of the earth with speed 5 m/s and the other fired from the surface of a planet with initial speed 3 m/s with the same angle of the projection, trace identical trajectories. Neglecting friction effect, the value of acceleration due to gravity on the planet is:

(1) 5.9 $\mathrm{m}/{\mathrm{s}}^2$

(2) 3.6 $\mathrm{m}/{\mathrm{s}}^2$

(3) 16.3 $\mathrm{m}/{\mathrm{s}}^2$

(4) 8.5 $\mathrm{m}/{\mathrm{s}}^2$

A projectile is projected from origin of a co-ordinate system with initial velocity of $2\hat{\mathrm{i}} + 3\hat{\mathrm{j}}$ m/s considering y-axis along vertical. The equation of its trajectory is (take g = 10 $\mathrm{m}/{\mathrm{s}}^2$)

(1) 6y = 4x - 5${\mathrm{x}}^2$

(2) 5y = 6x - 4${\mathrm{x}}^2$

(3) 4y = 6x - 5${\mathrm{x}}^2$

(4) 6y = 5x - 4${\mathrm{x}}^2$

During the downward motion of an oblique projectile from the top point of its trajectory, the radius of curvature of its trajectory will:

(1) Remain constant

(2) Increase

(3) Decrease

(4) May increase or decrease

A particle is thrown obliquely at \(t=0\). The particle has the same K.E. at \(t=5\) seconds and at \(t=9\) seconds. The particle attains maximum altitude at:
1. \(t=6\) s
2. \(t=7\) s
3. \(t=8\) s
4. \(t=14\) s

A particle of mass m is thrown at t=0 with kinetic energy ${\mathrm{K}}_0$ at an angle 30° with the horizontal from the top of a tower of height 80 m. The particle again has kinetic energy ${\mathrm{K}}_0$ after time ${\mathrm{t}}_0$. The ${\mathrm{t}}_0$ is equal to:

(1)  $\sqrt{\frac{2{\mathrm{K}}_0}{{\mathrm{mg}}^2}}$

(2)   $\sqrt{\frac{{\mathrm{K}}_0}{{\mathrm{mg}}^2}}$

(3)   $\sqrt{\frac{{\mathrm{K}}_0}{3{\mathrm{mg}}^2}}$

(4)   $\sqrt{\frac{{\mathrm{K}}_0}{4{\mathrm{mg}}^2}}$

The position vector of a particle is $\overrightarrow{\mathrm{r}}$ $=$ $\left(\mathrm{a}\right)$ $\sin$ $\mathrm{\omega t}\hat{\mathrm{i}}$ $+$ $\left(\mathrm{a}\right)$ $\cos$ $\mathrm{\omega t}\hat{\mathrm{j}}$. The velocity of the particle is:

Which of the following angles of projections will provide a maximum range to a projectile when projected with the same speed in all cases?

(1) 37°

(2) 54°

(3) 42°

(4) 49°

If a body is projected from the surface of the earth for maximum range R, then the maximum height attained by the body is:

(1) R

(2) $\frac{\mathrm{R}}{8}$

(3) 4R

(4) $\frac{\mathrm{R}}{4}$