Question

two bodies are projected with velocity u1 and u2 at angle theata1 and theata2 with respect to horizontal. what will be the nature of path of body A with respect to B? ​

Answer 1

Answer:

Given:-

Initial velocity of body A=$u_{1}$

Initial velocity of body B=$u_{2}$

Angle made by $u_{1}$ with the horizontal=$\theta_{1}$

Angle made by $u_{2}$ with the horizontal=$\theta_{2}$

To find:-

nature of path of body A with respect to B

solution:-

Initial displacement of body A = $X(0)_{1}$

Initial displacement of body B = $X(0)_{2}$

Initial displacement of body A with respect to B = $X(0)_{1}-X(0)_{2}$=0 m

Displacement of A at an instant t= Position vector of A at the instant = $X(t)_{1}$

Displacement of B at an instant t= Position vector of B at the instant = $X(t)_{2}$

Displacement of body A with respect to B at the instant t = $X(t)_{1}-X(t)_{2}$

In case of body A,

initial velocity=$u_{1}$

horizontal component of initial velocity=$u_{1}cos\theta _{1}$

vertical component of initial velocity=$u_{1}sin\theta _{1}$

Let the final velocity of body A be $v_{1}(u_{1}-gt)$

horizontal component of final velocity=$u_{1}cos\theta _{1}$(no acceleration)

vertical component of final velocity=$u_{1}sin\theta _{1}-gt$

we know that,

$v_{1}=(u_{1}cos\theta _{1})\hat{i}+(u_{1}sin\theta _{1}-gt)\hat{j}$

Similarly, for body B,

$v_{2}=(u_{2}cos\theta _{1})\hat{i}+(u_{2}sin\theta _{1}-gt)\hat{j}$

Relative velocity of A with respect to B=$v_{1}-v_{2}\\ =(u_{1}cos\theta _{1})\hat{i}+(u_{1}sin\theta _{1}-gt)\hat{j}-[(u_{2}cos\theta _{1})\hat{i}+(u_{2}sin\theta _{1}-gt)\hat{j}]\\ =(u_{1}cos\theta _{1}-u_{2}cos\theta _{2})\hat{i}+(u_{1}sin\theta _{1}-u_{2}sin\theta _{2})\hat{j}$

Relative path of A with respect to B=$X(t)_{1}-X(t)_{2}\\ =X(0)_{1}-X(0)_{2}+(v_{1}-v_{2})t\\ =0+[(u_{1}cos\theta _{1}-u_{2}cos\theta _{2})\hat{i}+(u_{1}sin\theta _{1}-u_{2}sin\theta _{2})\hat{j}]t\\ =(u_{1}cos\theta _{1}-u_{2}cos\theta _{2})t\: \hat{i}+(u_{1}sin\theta _{1}-u_{2}sin\theta _{2})t\: \hat{j}\_\_\_\_\_\_(ans)$

Answer 2

Answer:

Initial velocity of body A=

Initial velocity of body B=

Angle made by  with the horizontal=

Angle made by  with the horizontal=

To find:-

nature of path of body A with respect to B

solution:-

Initial displacement of body A =

Initial displacement of body B =

Initial displacement of body A with respect to B = =0 m

Displacement of A at an instant t= Position vector of A at the instant =

Displacement of B at an instant t= Position vector of B at the instant =

Displacement of body A with respect to B at the instant t =

In case of body A,

initial velocity=

horizontal component of initial velocity=

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Explanation: