PROJECTILE FIRED FROM AN INCLINED PLANE

 

Consider an inclined palne which makes an  with the horizontal. Let a projectile be projected with a velocity 'v' making an angle  with the horizontal.

 

 Let us choose X axis along the inclined plane and Y-axis perpendicular to it.

 

gcos and gsin are the two rectangular components of g.

 

Now,

 

v_x= v cos(-) and v_y = v sin(-).

 

Let T = time of flight of the projectile

 

The displacement of the projectile perpendicular to the inclined plane is clearly

 

Zero.

 

Using S = ut + (1/2) a t², for motion along y-axis, we get

 

0 = v sin(-) T -(1/2) g cos T²

 

or T = 2 v sin(-)/g cos  

 

Range on the inclined plane is

 

R = 2v² sin(-) cos/g cos²  

 

R_max= v²/g (1+sin)

 

H_max= v²sin²(-)/2g cos

 

If  is changed to -, then we get the formula of time of flight and range for

 

'down the inclined plane'. In this case,

 

T = 2v sin(+)/g cos  

 

R = 2v² sin(+) cos/g cos²

 

R_max= v²/g (1-sin)