Question

A body is projected with an initial velocity 20m/s at 60degree to the horizontal. its velocity after 1 second in the form of vectors is

Answer 1

Answer:

V = 10 i +  7.32 j

Given :

⇒ u = 20 m/s

⇒ angle with horizontal = α = 60°

To Find :

velocity after 1 second , V = ?

Solution :

At the place of projection ,

$u_{x} = 20 cos 60$

   = 20 × 1/2

   = 10 m/s

$u_{y} = 20sin60$

    = 20 × $\sqrt{3} /2$

    = 10$\sqrt{3}$ m /s

After 1 second ,

$v_{x} = 10m/s$  [ because x component of velocity remains same ]

$v_{y} = u_{y} + a_{y} .t$

$v_{y}$ =   10$\sqrt{3}$ - 10

$v_{y}$  =  10 ( $\sqrt{3}$ - 1 )

$v_{y}$   = 10 × 0.732

$v_{y}$   = 7.32 m/s

so , V in the form of vectors =

V = 10 i +  7.32 j

Answer 2

Answer:

V = 10 i + 7.32 j

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