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

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) $\frac{\mathrm{R}}{4}$

A projectile is projected from the ground with the velocity $v_{0}$ at an angle $\theta$ with the horizontal. What is the vertical component of the velocity of the projectile when its vertical displacement is equal to half of the maximum height attained?
1. $\sqrt{3} v_{0}\cos\theta$
2. $\frac{v_{0}}{\sqrt{2}} \sin\theta$
3. $\frac{v_{0}}{\sqrt{2}} \cos \theta$
4. $\sqrt{5} v_{0}$

A projectile is projected with a speed of 60 m/s at an angle of 60° from the horizontal. After some time projectile is moving at an angle of 30° with horizontal, what is the value of $\frac{\mathrm{d}}{\mathrm{dt}}\left|\overrightarrow{\mathrm{v}}\right|$ at this instant? (Given g = 10 $\mathrm{m}/{\mathrm{s}}^2$)

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

(2) $5\sqrt{3}$ $\mathrm{m}/{\mathrm{s}}^2$

(3) $\frac{5}{\sqrt{3}}$ $\mathrm{m}/{\mathrm{s}}^2$

(4) 5 $\mathrm{m}/{\mathrm{s}}^2$