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Then
0
  arc tan 1 45
2.The second level
What' sthe equation trajectory of body in Cartesian

coordinates?
The equation of uniform motion in horizontal projection (axis
OX) in Cartesian coordinates takes the form
x  v x t

Ditto the, equation of free fall accelerated motion in vertical

projection
( axis OY)

g t  2
y 
2
Hence there are two equations involving time t. To eliminate this time
we derive equation of trajectory
g 2
y  x
2v 2
x
i.e. the equation of parabola. Substitute numerical values and obtain
m
10
s 2 2 2
y  x  , 0 0125x
m 2
2  (.. 20 )
s 2
3.The third level

To define
- the normal and tangential acceleration of a body
-the radius of curvature of trajectory

In this case the total acceleration is the free fall acceleration (fig.1.13)
  
g  a  a
n

Or in scalar form
2
2
2
g  a  a .
n


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