A boy standing at the top of a tower of 20 m height drops a stone. Assuming g=$10 {\mathrm{ms}}^{-2}$, the velocity with which it hits the ground will be:
1. 20 m/s                                         
2. 40 m/s
3. 5 m/s                                           
4. 10 m/s

A body starts to fall freely under gravity. The distances covered by it in first, second and third second will be in the ratio: 
1. 1 : 3 : 5
2. 1 : 2 : 3
3. 1 : 4 : 9
4. 1 : 5 : 6

A body is thrown upwards and reaches its maximum height. At that position:

A ball is thrown vertically downwards with a velocity of \(20\) m/s from the top of a tower. It hits the ground after some time with the velocity of \(80\) m/s . The height of the tower is: (assuming $\mathrm{g}=10$ $\mathrm{m}/{\mathrm{s}}^2)$

A body starting from rest moves with uniform acceleration on a horizontal surface. The body covers 3 consecutive equal distances from the beginning in time ${\mathrm{t}}_1,$ ${\mathrm{t}}_2 , \mathrm{and}$ ${\mathrm{t}}_3$ seconds. The ratio of ${\mathrm{t}}_1:{\mathrm{t}}_2:{\mathrm{t}}_3$ is:
1. 1: 2 : 3
2. $1:\sqrt{2}:\sqrt{3}$
3. $1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$
4. $\sqrt{3}:\sqrt{2}:1$

A particle starts from rest (with constant acceleration) and acquires velocity 20 m/s in 5s. The distance travelled by the particle in the next 2 s will be:

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 student is standing at a distance of 50 metres from the bus. As soon as the bus begins its motion with an acceleration of 1 ms^–2, the student starts running towards the bus with a uniform velocity u. Assuming the motion to be along a straight road, the minimum value of u, so that the student is able to catch the bus is:

1. 5 ms^–1

2. 8 ms^–1

3. 10 ms^–1

4. 12 ms^–1

A particle is projected upwards. The times corresponding to height h while ascending and while descending are t₁ and t₂ respectively. The velocity of projection will be:

1. gt₁

2. gt₂

3. $g$ $(t_1+t_2)$

4. $\frac{g(t_1+t_2)}{2}$