No; horizontal range is maximum when 8 = 45° and maximum height attained by projectile is largest when \(\theta\) = 90°.

A vector cannot be zero if one of its components is non-zero.

Maximum horizontal distance = 100 m = \(\frac { \upsilon ^{ 2 } }{ g } \)
Maximum vertical height  = \(\frac { \upsilon ^{ 2 } }{ g } \)
=\(\frac{1}{2}\times\)100 = 50 m.

T₁=  \(\frac { 2u \ sin \ \theta }{ g } \) and T₂ = \(\frac { 2u\ sin(90^{ 0 }-\theta ) }{ g } \)
Now, \(\frac { { T }_{ 1 } }{ { T }_{ 2 } } =\frac { 2u \ sin \ \theta }{ g } \times \frac { g }{ 2u\ sin(90^{ 0 }-\theta ) } \)
= sin\(\theta\) cos \(\theta\) = tan \(\theta\)

When the two vectors are in mutually opposite directions.

No, since resultant of two vectors lie in the same plane.

Let u be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle of projection with horizontal, \(\theta ={ 45 }^{ o }\)  Then, \({ R }_{ max }={ u }^{ 2 }/g\)
Here,   \({ u }^{ 2 }/g=100m\)
In order to study the motion of the ball along vertical direction, consider a point on the surface of Earth as the origin and vetical upward direction as the positive direction of Y-axis. Taking motion of the ball along vertical upward direction, we have
\({ u }_{ y }=u,{ a }_{ y }=-g,{ v }_{ y }=0,t=?,{ y }_{ o }=0,y=?\)
As,  \({ v }_{ y }={ u }_{ y }+{ a }_{ y }t\)
\(\\ 0=u+(-g)t\Rightarrow t=u/g\)
Also,  \(y={ y }_{ o }+{ u }_{ y }t+\frac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }\)
\(\\ y=0+u(u/g)+\frac { 1 }{ 2 } (-g){ u }^{ 2 }/{ g }^{ 2 }\)
\(\\ =\frac { { u }^{ 2 } }{ g } -\frac { 1 }{ 2 } \frac { { u }^{ 2 } }{ g } =\frac { 1 }{ 2 } \frac { { u }^{ 2 } }{ g } =\frac { 100 }{ 2 } =50m\quad [\frac { { u }^{ 2 } }{ g } =100]\)

In the baseball game, a player has to throw a ball so that it goes a certain distance in the minimum time. The time would depends on velocity of ball and angle of throw with the horizontal. Thus, while playing a baseball game, the speed and angle of projection have to be adjusted suitable so that the ball covers the desired distance in minimum time. So, a player has to see the distance and air resistance while playing with a baseball game.

When a stone is dropped from a railway carriage, it will fall vertically downwards with acceleration due to gravity g. Therefore, with respect to earth, the acceleration of the stone will be g only, Inside the carriage the stone has two accelerations: (i) horizontal acceleration a, due to the motion of the carriage and (ii) vertical acceleration due to gravity g. Thus, the acceleration of the stone with respect to carriage is \(\sqrt { { a }^{ 2 }+{ g }^{ 2 } } \)

(i) The nature of the path will be vertical straight line because the ball has only one velocity acting vertically. The horizontal component of velocity of ball will not be visible to the passenger inside the train because he himself is in motion.
(ii) Tile nature of the path will be a parabolic path because the ball has the vertical as well as horizontal component of velocities.

\(\therefore \)   Ratio is 
\(\frac { A.B }{ \left| A.B \right| } =\frac { AB\quad cos\quad \theta }{ AB\quad sin\quad \theta } =cot\quad \theta \)
\(\\ =cot\quad { 60 }^{ o }=\frac { 1 }{ \sqrt { 3 } } \)
\(\\ As,\quad\theta ={ 60 }^{ o },cot{ 60 }^{ o }=\frac { 1 }{ \sqrt { 3 } } \) 

We know that time always flows on and on i.e. from the past to present and then to future.
Therefore, a direction can be assigned to time. Since the direction of time is unique and it is unspecified or unstated. That is why, time cannot be a vector though it has a direction.