A particle is projected from a horizontal plane (x-z plane) such that its velocity vector at time t is given by \(\overrightarrow{\mathrm{v}}=a \hat{i}+(b-c t )\hat{j}.\) Its range on this horizontal plane is given by: 

1.	\(\frac{\mathrm{ba}}{\mathrm{c}} \)	2.	\(\frac{2 \mathrm{ba}}{\mathrm{c}} \)
3.	\(\frac{3 \mathrm{ba}}{\mathrm{c}} \)	4.	None

A river is 1 km wide. The banks are straight and parallel. The current is 5 km/h and is parallel to the banks. A boat has a maximum speed of 3 km/h in still water. In what direction should the boat head so as to arrive at point B directly opposite to its starting point A?

1. directly across the river

2. head $53°$ upstream from the line AB

3. head $37°$ $\mathrm{upstream}$ $\mathrm{from}$ $\mathrm{the}$ $\mathrm{line}$ $\mathrm{AB}$

4. the trip from A to B is not possible with this speed

A particle starting from the point \((1,2)\) moves in a straight line in the XY-plane. Its coordinates at a later time are \((2,3).\) The path of the particle makes with x-axis an angle of:

A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity)

A man moving in the west direction observes wind is blowing towards the south. If the man doubles his speed in the same direction, then the direction of wind with respect to man will be:

1.  North-West

2.  South-West

3.  North-East

4.  East-South

Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings (building being \(100~\text{m}\) apart and being of the same height of \(200~\text{m}\)) with the same velocity of \(25~\text{m/s}\). When and where will the two bullets collide? \(\left(g = 10~\text{m/s}^2 \right)\)?

A man standing on a road holds his umbrella at 30° with the vertical to keep the rain away. He throws the umbrella and starts running at 10 km/hr. He finds that raindrops are hitting his head vertically. The speed of raindrops with respect to the road will be

1. 10 km/hr

2. 20 km/hr

3. 30 km/hr

4. 40 km/hr

Two particles A and B, move with constant velocities \(\vec{v_1}\)   and  \(\vec{v_2}\) . At the initial moment their position vector are  \(\vec{r_1}\) and  \(\vec{r_2}\)  respectively. The condition for particles A and B for their collision to happen will be:

1.  $\overrightarrow{{\mathrm{r}}_{1 }}.\overrightarrow{{\mathrm{v}}_1} =\overrightarrow{{\mathrm{r}}_{2 }}.\overrightarrow{{\mathrm{v}}_2}$

2.  $\overrightarrow{{\mathrm{r}}_1} \mathrm{x} \overrightarrow{{\mathrm{v}}_1} =\overrightarrow{{\mathrm{r}}_2} \mathrm{x} \overrightarrow{{\mathrm{v}}_2}$

3. $\overrightarrow{{\mathrm{r}}_1}- \stackrel{\rightharpoonup }{{\mathrm{r}}_2} =\overrightarrow{{\mathrm{v}}_1}- \overrightarrow{{\mathrm{v}}_2}$

4. $\frac{\overrightarrow{{\mathrm{r}}_1}- \overrightarrow{{\mathrm{r}}_2}}{\left|\overrightarrow{{\mathrm{r}}_1}- \overrightarrow{{\mathrm{r}}_2}\right|}= \frac{\overrightarrow{{\mathrm{v}}_2}- \overrightarrow{{\mathrm{v}}_1}}{\left|\overrightarrow{{\mathrm{v}}_2}- \overrightarrow{{\mathrm{v}}_1}\right|}$

A particle moves along a parabolic path y = 9x² in such a way that the x component of the velocity remains constant and has a value of $\frac{1}{3}$ ${\mathrm{ms}}^{-1}$. It can be deduced that the acceleration of the particle will be:

1. $\frac{1}{3}$ $\hat{\mathrm{j}}$ ${\mathrm{ms}}^{-2}$

2. $3$ $\hat{\mathrm{j}}$ ${\mathrm{ms}}^{-2}$

3. $\frac{2}{3}$ $\hat{\mathrm{j}}$ ${\mathrm{ms}}^{-2}$

4. $2$ $\hat{\mathrm{j}}$ ${\mathrm{ms}}^{-2}$