A ball is thrown vertically upward and it reaches the highest point in \(4~\text s.\) Immediately, a second ball is thrown upwards with an initial speed that is twice that of the first. The second ball meets the first after a time:
1. \(1~\text s\) 
2. \(2~\text s\) 
3. \(3~\text s\)
4. \(4~\text s\)

A particle projected vertically under gravity passes a certain level on the way up at a time \(T_1\) and on the way down at a time \(T_2\) – after it was projected. The speed of projection is:

1. \(\dfrac{1}{2} g\left(T_{1}+T_{2}\right)\)

2. \(\dfrac{1}{2} g\left(T_{1}-T_{2}\right)\)

3. \(g \sqrt{T_{1} T_{2}}\)

4. \(\dfrac{1}{2} g \dfrac{T_{1} T_{2}}{T_{1}+T_{2}}\)

A boy throws a ball straight up the side of a building and receives it after \(4~\text s.\) On the other hand, if he throws it so that it strikes a ledge on its way up, it returns to him after \(3~\text s.\) The ledge is at a distance \(d\) below the highest point, where \(d=? \) (take acceleration due to gravity, \(g=10~\text{ms}^{-2})\)
1. \(5~\text m\)
2. \(2.5~\text m\)
3. \(1.25~\text m\)
4. \(10~\text m\)

A man driving a scooter at \(15~\text{m/s}\) brakes at the rate of \(2~\text{m/s}^2\). His speed, after \(10~\text{s}\) after the application of brakes will be:
1. \(5~\text{m/s}\)
2. \(-5~\text{m/s}\)
3. \(0~\text{m/s}\)
4. \(10~\text{m/s}\)

A man \((A)\) has to throw a ball vertically up to a partner \((B)\) who is standing up, above his level by \(15~\text{m}.\) The \((B)\) partner can catch the ball only when it comes downwards with a maximum speed of \(10~\text{m/s}\)
(take acceleration due to gravity as \(10~\text{m/s}^{2}\))
The minimum and maximum speeds of the throw are: (nearly)
1. \(10~\text{m/s}~\text{and}~20~\text{m/s}\)
2. \(10~\text{m/s}~\text{and}~30~\text{m/s}\)
3. \(20~\text{m/s}~\text{and}~20\sqrt{3}~\text{m/s}\)
4. \(10\sqrt{3}~\text{m/s}~\text{and}~20~\text{m/s}\)