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

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 \(t_1\) and \(t_2\) 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 \(1~\text{m}\) each will then be:

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) $ut$