A boy standing at the top of a tower of \(20\) m height drops a stone. Assuming \(g=10\) m/s², 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 the first, second and third second will be in the ratio: 

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 \(g = 10~\text{m/s}^2)\)$$

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 \(t_1, t_2,\text{and}~t_3\)${\mathrm{}}_{}$ seconds. The ratio of \(t_1:t_2:t_3\) is:
1. \(1:2:3\)
2. \(1:\sqrt{2}:\sqrt{3}\)
3. \(1:\left(\sqrt{2}-1\right):\left(\sqrt{3}-\sqrt{2}\right)\)
4. \(\sqrt{3}:\sqrt{2}:1\)$\mathrm{}$

A particle starts from rest (with constant acceleration) and acquires velocity \(20\) m/s in \(5\) s. 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 \(t_1\) 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 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. \(gt_1\)
2. \(gt_2\)
3. \(g(t_1+t_2)\)
4. \(\frac{g(t_1+t_2)}{2}\)