A packet is dropped from a balloon which is going upwards with the velocity 12 m/s, the velocity of the packet after 2 seconds will be 

1. –12 m/s

2. 12 m/s

3. –7.6 m/s

4. 7.6 m/s

If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first three seconds, the time of the travel is:
1. $6$ sec
2. $5$ sec
3. $4$ sec
4. $3$ sec

The effective acceleration of a body, when thrown upwards with acceleration a will be: 

1. $\sqrt{a-g^2}$

2. $\sqrt{a^2+g^2}$

3. $(a-g)$

4. $(a+g)$

A body is thrown vertically upwards with velocity $u.$ The distance travelled by it in the fifth and the sixth seconds are equal. The velocity $u$ is given by ($g = 9.8~\text{m/s}^2$) 

1. $24.5~\text{m/s}$
2. $49.0~\text{m/s}$
3. $73.5~\text{m/s}$
4. $98.0~\text{m/s}$

A parachutist after bailing out falls $50~\text{m}$ without friction. When parachute opens, it decelerates at $2~\text{m/s}^2$. He reaches the ground with a speed of $3~\text{m/s}$. At what height, did he bail out ?
1. $293~\text{m}$
2. $111~\text{m}$
3. $91~\text{m}$
4. $182~\text{m}$

When a ball is thrown up vertically with velocity v₀, it reaches a maximum height of 'h'. If one wishes to triple the maximum height then the ball should be thrown with velocity

1. $\sqrt{3}{v}_o$

2. 3v₀

3. 9v₀

4. (3/2)v₀

A particle moving in a straight line covers half the distance with a speed of $3~\text{m/s}$. The other half of the distance is covered in two equal time intervals with speeds of $4.5~\text{m/s}$ and $7.5~\text{m/s}$ respectively. The average speed of the particle during this motion is:
1. $4.0~\text{m/s}$
2. $5.0~\text{m/s}$
3. $5.5~\text{m/s}$
4. $4.8~\text{m/s}$

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity of v₀. The distance travelled by the particle in time t will be:

1. $v_{0}t+\frac{1}{3}b{t}^2$

2. $v_{0}t+\frac{1}{3}b{t}^3$

3. $v_{0}t+\frac{1}{6}b{t}^3$

4. $v_{0}t+\frac{1}{2}b{t}^2$

A particle starts from rest. Its acceleration $(a)$ versus time $(t)$ is as shown in the figure. The maximum speed of the particle will be: 

             
1. $110~\text{m/s}$
2. $55~\text{m/s}$
3. $550~\text{m/s}$
4. $660~\text{m/s}$

A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is

1.  $\left(\frac{{\displaystyle {\alpha }^2+{\beta }^2}}{{\displaystyle \alpha \beta }}\right)t$
2.  $\left(\frac{{\displaystyle {\alpha }^2-{\beta }^2}}{{\displaystyle \alpha \beta }}\right)\hspace{0.17em}t$
3.  $\frac{(\alpha +\beta )t}{\alpha \beta }$
4.  $\frac{\alpha \beta t}{\alpha +\beta }$