A projectile is projected at an angle of 60° from horizontal. At an instant, it is moving at an angle of 30° with horizontal with velocity 10 m/s. The radial acceleration of the particle at that instant is:

1.  $\frac{20}{\sqrt{2}}\mathrm{m}/{\mathrm{s}}^2$

2.  $\frac{20}{\sqrt{3}}\mathrm{m}/{\mathrm{s}}^2$

3.  $20\sqrt{3}\mathrm{m}/{\mathrm{s}}^2$

4.  $5\sqrt{3}\mathrm{m}/{\mathrm{s}}^2$

The initial velocity of a projectile is given by $\mathrm{v} = \left(20\hat{\mathrm{i}} + 50\hat{\mathrm{j}} + 70\hat{\mathrm{k}}\right)$ m/s. If acceleration due to gravity is g = 10($-\hat{\mathrm{k}}$) m/${\mathrm{s}}^2$, then the time after which projectile returns to the same horizontal level is:

(1) 4 s

(2) 10 s

(3) 14 s

(4) 12 s

Two persons start moving along two crossroads with the same speed 5 m/s. One in the direction of north and other is in the direction of the east as shown in the figure. Angular velocity of A with respect to B is :

(1) 0.5 m/s

(2) Zero

(3) 1.0 m/s

(4) 2 m/s

In projectile motion, accelerations of the projectile when it is gaining height and losing height respectively are

(1) g upward, g upward

(2) g upward, g downward

(3) g downward, g downward

(4) g downward, g upward

A person, who can swim with speed \(u\) relative to water, wants to cross a river (of width \(d\) and water is flowing with speed \(v\)). The minimum time in which the person can do so is:
1. \(\frac{d}{v}\)
2. \(\frac{d}{u}\)
3. \(\frac{d}{\sqrt{v^{2}   +   u^{2}}}\)
4. \(\frac{d}{\sqrt{v^{2}   -   u^{2}}}\)

The position vector of a particle \(\overrightarrow r\) as a function of time \(t\) (in seconds) is \(\overrightarrow r=3 t \hat{i}+2t^2\hat j~\text{m}\). The initial acceleration of the particle is:
1. \(2~\text{m/s}^2\)
2. \(3~\text{m/s}^2\)
3. \(4~\text{m/s}^2\)
4. zero

A particle is projected on a horizontal surface with kinetic energy K at an angle $\theta$ with the vertical. The kinetic energy of the particle at the highest point of trajectory is:

(1) K

(2) Zero

(3) K ${\cos }^2\mathrm{\theta }$

(4) K ${\sin }^2\mathrm{\theta }$

Velocity and acceleration vectors of a particle moving on a circular path are shown here. At the instant shown:

(1) The speed of the particle must be decreasing.

(2) Acceleration of particle must be increasing in magnitude.

(3) The dot product of $\overrightarrow{\mathrm{v}}$ and $\overrightarrow{\mathrm{a}}$ is positive.

(4) The magnitude of the momentum of the particle must be increasing in magnitude.

A cricketer can throw a ball to a maximum horizontal distance of 50 m. How much high above the ground can he throw the same ball?

(1) 50 m

(2) 25 m

(3) 75 m

(4) 100 m

Coordinates of a particle as a function of time \(t\) are \(x= 2t\),
\(y =4t\). It can be inferred that the path of the particle will be: