A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds? 

1. h/9 meters from the ground

2. 7h/9 meters from the ground

3. 8h/9 meters from the ground

4. 17h/18 meters from the ground

Two balls of same size but the density of one is greater than that of the other are dropped from the same height, then which ball will reach the earth first (air resistance is negligible) 

(1) Heavy ball

(2) Light ball

(3) Both simultaneously

(4) Will depend upon the density of the balls

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) $v_{0}t+\frac{1}{2}b{t}^2$