A ball is thrown vertically downwards with a velocity of \(20\) m/s from the top of a tower. It hits the ground after some time with the velocity of \(80\) m/s . The height of the tower is: (assuming $\mathrm{g}=10$ $\mathrm{m}/{\mathrm{s}}^2)$

1.	\(340\) m	2.	\(320\) m
3.	\(300\) m	4.	\(360\) m

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 person travelling in a straight line moves with a constant velocity \(v_1\) for a certain distance \(x\) and with a constant velocity \(v_2\) ${\mathrm{}}_{}$for the next equal distance. The average velocity \(v\) is given by the relation:

A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time $t_1$ if the elevator is moving uniformly and time $t_2$ if the elevator is stationary. Then:

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