Physics, asked by COOLPJ3582, 11 months ago

A particle is projected from groundwith speed 80m/s at an angle 30 dege

Answers

Answered by UrvashiBaliyan
10

Explanation:

The time of flight = .

So at 4 seconds the body is at the higest point.

At t=2s and t= 6 second the body is at the same height. So Vertical displacement between 2s and 6 s is zero.( Body went up and returned to the same vertical level.

So vertical average velocity for this period is zero.

Horizontal velocity during this period is constant which is equal to .

This is the average velocity between t = 2 s and t = 6 s.

Answered by nrathour769
1

Let's see the time taken by the particle to reach maximum height,it is,

t

=

u

sin

θ

g

Given,

u

=

80

m

s

1

,

θ

=

30

so,

t

=

4.07

s

That means at

6

s

it already started moving down.

So,upward displacement in

2

s

is,

s

=

(

u

sin

θ

)

2

1

2

g

(

2

)

2

=

60.4

m

and displacement in

6

s

is

s

=

(

u

sin

θ

)

6

1

2

g

(

6

)

2

=

63.6

m

So,vertical dispacement in

(

6

2

)

=

4

s

is

(

63.6

60.4

)

=

3.2

m

And horizontal displacement in

(

6

2

)

=

4

s

is

(

u

cos

θ

4

)

=

277.13

m

So,net displacement is

4

s

is

3.2

2

+

277.13

2

=

277.15

m

So,average velcoity = total displacement /total time=

277.15

4

=

69.29

m

s

1

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