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:

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.  $\frac{\alpha \beta t}{\alpha +\beta }$