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 $\mathrm{m}/{\mathrm{s}}^2$

A particle is rotating with increasing speed on a circular track. The angle between its radius vector and acceleration is $\theta$, then:

(1) 0° < $\mathrm{\theta }$ < 90°

(2) 45° < $\mathrm{\theta }$ < 90°

(3) 90° < $\mathrm{\theta }$ < 180°

(4) $\mathrm{\theta }$ = 90°

A particle starts moving on a circular path from rest, such that its tangential acceleration varies with time as \(a_t=kt\). Distance traveled by particle on the circular path in time \(t\) is:
1. \( \frac{kt^3}{3} \)
2. \(\frac{kt^2}{6} \)
3. \(\frac{kt^3}{6} \)
4. \(\frac{k t^2}{2}\)

A projectile thrown from the ground has horizontal range R. If velocity at the highest point is doubled somehow, the new range will be:

(1) 3 R

(2) 1.5 R

(3) $\sqrt{3}$ R

(4) 2 R

The time of flights for a particle thrown for equal range at different angles is 4 sec and 3 sec. Speed of projection is:

(1) 10 m/s

(2) 20 m/s

(3) 25 m/s

(4) 50 m/s

A body is projected from the ground at a speed v at an angle $\mathrm{\theta }$ above the horizontal. The radius of curvature of its path at the highest point is:

(1)  $\frac{{\mathrm{v}}^2<mspace\; width="1em"></mspace>{\cos }^2<mspace\; width="1em"></mspace>\mathrm{\theta }}{2\mathrm{g}}$

(2)  $\frac{\mathrm{v}<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>\mathrm{\theta }}{\mathrm{g}}$

(3)  $\frac{\mathrm{v}<mspace\; width="1em"></mspace>{\cos }^2<mspace\; width="1em"></mspace>\mathrm{\theta }}{\mathrm{g}}$

(4)  $\frac{{\mathrm{v}}^2<mspace\; width="1em"></mspace>{\cos }^2<mspace\; width="1em"></mspace>\mathrm{\theta }}{\mathrm{g}}$

A particle is projected from the origin with velocity $\left(4\hat{\mathrm{i}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>9\hat{\mathrm{j}}\right)$ m/s. The acceleration in the region is constant and -10$\hat{\mathrm{j}}$ $\mathrm{m}/{\mathrm{s}}^2$. The magnitude of velocity after one second is

(1) 8 m/s

(2) $\sqrt{15}$ m/s

(3) $\sqrt{17}$ m/s

(4) $\sqrt{27}$ m/s