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 speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is:  

1. 4.0 m/s

2. 5.0 m/s

3. 5.5 m/s

4. 4.8 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 m/s

2. 55 m/s

3. 550 m/s

4. 660 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.  $\frac{\alpha \beta t}{\alpha +\beta }$

A stone dropped from a building of height h and reaches the earth after t seconds. From the same building, if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t₁ and t₂ seconds respectively, then: 

1. $t=t_1-t_2$

2. $t=\frac{t_1+t_2}{2}$

3. $t=\sqrt{t_1{t}_2}$

4. $t=t_1^2{t}_2^2$ 

A ball is projected upwards from a height h above the surface of the earth with velocity v. The time at which the ball strikes the ground is

(1) $\frac{v}{g}+\frac{2hg}{\sqrt{2}}$

(2) $\frac{v}{g}\left[1-\sqrt{1+\frac{2h}{g}}\right]$

(3) $\frac{v}{g}\left[1+\sqrt{1+\frac{2gh}{v^2}}\right]$

(4) $\frac{v}{g}\left[1+\sqrt{v^2+\frac{2g}{h}}\right]$

A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of 1 m each will then be:
1. All equal, being equal to \(\sqrt{2 / g} \) second. 
2. In the ratio of the square roots of the integers 1, 2, 3.....
3.  In the ratio of the difference in the square roots of the integers \(\sqrt{1}\), \((\sqrt{2}-\sqrt{1})\),\((\sqrt{3}-\sqrt{2})\),\((\sqrt{4}-\sqrt{3})\) \( \ldots\) 
4.  In the ratio of the reciprocal of the square roots of the integers i.e... \(\frac{1}{\sqrt{1}}\), \(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{3}}\),\(\frac{1}{\sqrt{4}} \) 

A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time (Given $g=9.8m/s^2)$

(1) At least 0.8 m/s

(2) Any speed less than 19.6 m/s

(3) Only with speed 19.6 m/s

(4) More than 19.6 m/s

If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is 

(1) $\frac{1}{2}g{t}^2$

(2) $ut-\frac{1}{2}g{t}^2$

(3) $(u-gt)t$

(4) $ut$