A person sitting on the ground floor of a building notices through the window, of height \(1.5~\text{m}\), a ball dropped from the roof of the building crosses the window in \(0.1~\text{s}\). What is the velocity of the ball when it is at the topmost point of the window? \(\left(g = 10~\text{m/s}^2\right )\)
1. \(15.5~\text{m/s}\)
2. \(14.5~\text{m/s}\)
3. \(4.5~\text{m/s}\) 
4. \(20~\text{m/s}\) 

A particle covers half of its total distance with speed ν₁ and the rest half distance with speed ν₂.
Its average speed during the complete journey is:

1. $\frac{v_1+v_2}{2}$

2. $\frac{v_1{v}_2}{v_1+v_2}$

3. $\frac{2v_1{v}_2}{v_1+v_2}$

4. $\frac{v_1^2{v}_2^2}{v_1^2+v_2^2}$

The motion of a particle is given by the equation $S=(3t^3+7t^2+14t+8)m,$ The value of the acceleration of the particle at t = 1 s is :

The displacement \(x\) of a particle varies with time \(t\) as \(x = ae^{-\alpha t}+ be^{\beta t}\)$\mathrm{}$
, where \(a,\) \(b,\) \(\alpha,\) and \(\beta\) are positive constants. The velocity of the particle will:

A car is moving with velocity v. It stops after applying breaks at a distance of 20 m. If the velocity of the car is doubled, then how much distance it will cover (travel) after applying breaks?
1.  40 m
2.  80 m
3.  160 m
4.  320 m

A body starts falling from height 'h' and if it travels a distance of h/2 during the last second of motion, then the time of flight is (in seconds):

1. $\sqrt{2}$ $-$ $1$

2. $2$ $+$ $\sqrt{2}$ $$

3. $\sqrt{2}$ $$ $+$ $\sqrt{3}$

4. $\sqrt{3}$ $$ $+$ $2$

For a particle, displacement time relation is given by $t$ $=$ $\sqrt{x}$ $+$ $3$ . Its displacement, when its velocity is zero will be:
1. \(2\) m
2. \(4\) m
3. \(0\)
4. none of the above

A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 feet tall building. After a fall of 30 feet each towards the earth, their respective kinetic energies will be in the ratio of:

1. 1: 4

2. 1: 2

3. 1: $\sqrt{2}$

4. $\sqrt{2}$ :1

A particle starts from rest with constant acceleration. The ratio of space-average velocity to the time-average velocity is:

where time-average velocity and space-average velocity, respectively, are defined as follows: 

$<v{>}_{time}$ $=$ $\frac{\int vdt}{\int dt}$
$<v{>}_{space}$ $=$ $\frac{\int vds}{\int ds}$ $$

1. $\frac{1}{2}$

2. $\frac{3}{4}$

3. $\frac{4}{3}$

4. $\frac{3}{2}$

A particle is thrown vertically upward. Its velocity at half its height is \(10\) m/s. Then the maximum height attained by it is: (Assume, \(g=\) \(10\) m/s²) 
1. \(8\) m
2. \(20\) m
3. \(10\) m
4. \(16\) m