Jimenez

answers: 1

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Answer:

ANSWER

For particle projected up the plane, in this situation we take two axes , one perpendicular to the plane and other along the plane then , taking component of velocity and acceleration along these axes we get :

along x-axis u

x

=10cos30

0

=5

3

,a

x

=−gsin30

0

=−5

and along y-axis, u

y

=10sin30

0

=5,a

y

=−gcos30

0

=−5

3

using these we get, time of flight T=

a

y

2u

y

=

5

3

2×5

=

3

2

so, v

x

=u

x

+a

x

t=5

3

−5×

3

2

=

3

5

3

m/s

and v

y

=u

y

+a

y

t=5−5

3

×

3

2

=−5m/s

so speed with which particle strike the inclined will be v=

(

v

x

2

+v

x

2

)=

(

25/3+25)=

3

10

m/s

and component of velocity perpendicular to the plane when it strikes at A is 5m/s

467

Lyuba

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