Physics, asked by beastmuze, 11 months ago

A body of mass m is projected from ground with
speed u at an angle Theta with horizontal then find
aut :-
1) Initial velocity vector
2) Velocity vector at topmost point
3) velocity vector when body Strikes on ground
4) Change in momentum between point of projection and striking point
on ground.​

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Answers

Answered by ranjanalok961
1

Explanation:

A body of mass m is projected from ground with

A body of mass m is projected from ground withspeed u at an angle Theta with horizontal then

Let theta = x,

1) Initial velocity vector = u sinx i + u cosx j

2) Velocity vector at topmost point = 0 i + u cos j

3) velocity vector when body Strikes on ground = u sinx i + u cosx (-j)

4) Change in momentum between point of projection and striking point

4) Change in momentum between point of projection and striking pointon ground = m ( v final -v initial) =| m( 2 u cosx) |

Answered by Anonymous
0

initial speed = u

angle = theta

1)

We know

x component of a vector U= ucos\thetaθ

y component of vector U= usin\thetaθ

Uvector= \pink{(u \cos\theta \: ) i + (u \sin \theta )j}(ucosθ)i+(usinθ)j -----i

━━━━━━━━━━━━━━━

2.

Now clearly from the diagram,

Vvector=

\pink{(u \: cos \theta) i - (u \: sin \: \theta) j}(ucosθ)i−(usinθ)j ----ii

━━━━━━━━━━━━━━━

3.

At the top,

the vertical velocity will be zero or Uy=0

thus at top,

Vvector

= (Ux)i +(Uy)j

=(Ux)i

=\blue{(u \: cos \: \theta) i}(ucosθ)i

━━━━━━━━━━━━━━━━

4.

Momentum:

=m (Vfinal- Vinitial) [from equation i and ii]

=\red{m( - 2u \: sin \: \theta) j}m(−2usinθ)j

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