A ball of mass m₁ and another ball of mass m₂ are dropped from equal height. If time taken by the balls are t₁ and t₂ respectively, then 

1. $t_1=\frac{t_2}{2}$

2. $t_1=t_2$

3. $t_1=4t_2$

4. $t_1=\frac{t_2}{4}$ 

With what velocity a ball be projected vertically so that the distance covered by it in 5^th second is twice the distance it covers in its 6^th second $(g=10m/s^2)$ 

(1) 58.8 m/s

(2) 49 m/s

(3) 65 m/s

(4) 19.6 m/s

A ball is dropped downwards. After 1 second another ball is dropped downwards from the same point. What is the distance between them after 3 seconds 

1.  25 m
2.  20 m
3.  50 m
4.  9.8 m

A stone is thrown with an initial speed of 4.9 m/s from a bridge in vertically upward direction. It falls down in water after 2 sec. The height of the bridge is 

(1) 4.9 m

(2) 9.8 m

(3) 19.8 m

(4) 24.7 m

A stone is shot straight upward with a speed of 20 m/sec from a tower 200 m high. The speed with which it strikes the ground is approximately 

(1) 60 m/sec

(2) 65 m/sec

(3) 70 m/sec

(4) 75 m/sec

A body freely falling from the rest has a velocity ‘v’ after it falls through a height ‘h’. The distance it has to fall down for its velocity to become double, is 

(1) 2 h

(2) 4 h

(3) 6 h

(4) 8 h

Velocity of a body on reaching the point from which it was projected upwards, is 

1. $v=0$

2. $v=2u$

3. $v=0.5u$

4. $v=-u$

A body projected vertically upwards with a velocity u returns to the starting point in 4 seconds. If g = 10m/sec², the value of u is  

(1) 5 m/sec

(2) 10 m/sec

(3) 15 m/sec

(4) 20 m/sec

Time taken by an object falling from rest to cover the height of h₁ and h₂ is respectively t₁ and t₂ . Then the ratio of t₁ to t₂ is 

(1) h₁ : h₂

(2) $\sqrt{h_1}:\sqrt{h_2}$

(3) h₁ : 2h₂

(4) 2h₁ : h₂

A body is thrown vertically up from the ground. It reaches a maximum height of \(100\) m in \(5\) s. After what time will it reach the ground from the position of maximum height?