The gravity in space is given by $\overrightarrow{\mathrm{g}}=-10 \hat{j} m{s}^{-2}$. Two particles are simultaneously projected with velocity $\overrightarrow{{\mathrm{u}}_1}=\left(10 m{s}^{-1}\right)\hat{i}+\left(10 m{s}^{-1}\right)\hat{j}$ and $\overrightarrow{{\mathrm{u}}_2}=\left(20 m{s}^{-1}\right)\hat{i}+\left(10 m{s}^{-1}\right)\hat{j}$. Then, the ratio of their times of flight

1. 1:1

2. 1:2

3. 2:1

4. none

What determines the nature of the path followed by the particle?

(1) Speed only

(2) Velocity only

(3) Acceleration only

(4) None of these

A boat is sent across a river in perpendicular direction with a velocity of 8 km/hr. If the resultant velocity of boat is 10 km/hr, then velocity of the river is : 

(1) 10 km/hr

(2) 8 km/hr

(3) 6 km/hr

(4) 4 km/hr

A river is flowing from W to E with a speed of 5 m/min. A man can swim in still water with a velocity 10 m/min. In which direction should the man swim so as to take the shortest possible path to go to the south. 

(1) 30° with downstream

(2) 60° with downstream

(3) 120° with downstream

(4) South

A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is 

(1) East-north direction

(2) West-north direction

(3) South-east direction

(4) None of these

A ball P is dropped vertically and another ball Q is thrown horizontally from the  same height and at the same time. If air resistance is neglected, then 

(1) Ball P reaches the ground first

(2) Ball Q reaches the ground first

(3) Both reach the ground at the same time

(4) The respective masses of the two balls will decide the time

A frictionless wire AB is fixed on a sphere of radius R. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to B is:
          

1. $\frac{2\sqrt{gR}}{g\cos \theta }$

2. $2\sqrt{gR}.\frac{\cos \theta }{g}$

3. $2\sqrt{\frac{R}{g}}$ 

4. $\frac{gR}{\sqrt{g\cos \theta }}$

A body is slipping from an inclined plane of height h and length l. If the angle of inclination is θ, the time taken by the body to come from the top to the bottom of this inclined plane is:

1. $\sqrt{\frac{2h}{g}}$

2. $\sqrt{\frac{2l}{g}}$

3. $\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}$ 

4. $\sin \theta \sqrt{\frac{2h}{g}}$

An aeroplane is moving with a velocity \(u\). It drops a packet from a height \(h\). The time \(t\) taken by the packet in reaching the ground will be:
1. \( \sqrt{\left(\frac{2 g}{h}\right)} \)
2. \( \sqrt{\left(\frac{2 u}{g}\right)} \)
3. \( \sqrt{\left(\frac{h}{2 g}\right)} \)
4. \( \sqrt{\left(\frac{2 h}{g}\right)}\)

An aeroplane is moving with horizontal velocity u at height h. The velocity of a packet dropped from it on the earth's surface will be (g is acceleration due to gravity) 

(1) $\sqrt{u^2+2gh}$

(2) $\sqrt{2gh}$

(3) 2 gh

(4) $\sqrt{u^2-2gh}$