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2
gt 2 h
h   t 
2 g

m
2  20
s
so that t   s 2
m
10
s 2

-uniform motion in horizontal direction with speed v
x
S  v x t
m
Therefore 20  s 2  40 m
s

Velocity of curvilinear motion of material point as vector is tangent to
the trajectory ,then as follows from fig.1.12

2
2
v  v  v
x y
where v is speed of the free falling.Taking to account, that
y
m m
,v  g  t  10 s 2  20
y
s 2 s
we obtain

m
2
2
v  v  ( gt)  28 3 , .
x
s

As follows from the right triangle (fig.1.12 ) angle between
velocity and horizon at the moment of falling is defined by the next
trigonometrical relationship

v y
tg  .
v x

Therefore
m
20
v y s
tg    1
v x 20 m

s

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