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An arrow shot from a bow is an example of projectile motion a)How does horizontal and vertical component in a projectile motion b)Obtain expression for the time of flight, horizontal range and maximum height for a particle with initial velocity u making angle teta with the horizontal?
A ) A projectile's total velocity can be split into two components as  horizontal and vertical velocity. Consider  the figure 
                                                                  
u = initial velocity  
x = angle of projection  (degrees) 
uh = horizontal velocity  
uv = initial vertical velocity  

The  object is fired at an angle of ‘x’ with total velocity ‘u’,from the basis of trigonometry horizontal and vertical components can be derived as follows :   
uh / u = sinx
      uh = u sinx  (m/s)   &
uv / u = cosx
       uv = ucosx (m/s) 

B ) Maximum Projectile Range

        d = (V2sin(2?)) / g

 Let the  value of 2? for which sin 2? = 1 is 90?. ie, the value of ? = 90/2 = 45?

so,     d =V2/g

Maximum height covered by the projectile :

hmax = uv2 /19.6 

uv -  initial vertical velocity

The time of flight of projectile is :

     T= 2Usin(?) / g 


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