Question

Two particles 'A' and 'B' are projected in the vertical plane with same initial speed u, from part (0,0) and (l,-h) towards each other as shown in figure at t= 0.

I) The path of particle A wrt B will be

II) Minimum distance between particle A and B during this motion

III) The time when seperation between A and B is minimum is ?​

Answer 1

Explanation:

1) VAx=u0cos theta

VBx=-u0cos theta

VABx=u0cos theta-gt

VBY= u0 sin theta -gt

i.e VABY=0

SO WRT B A IS moving in the straight line as it has only constant velocity VABX=2U0COS THETA along x - axis

2) at any time t

xA=u0cos theta

xb=l-u cos theta-t

ya=usin theta t-1/2 gt^2

yb= -d(usin thetat-1/2 gt^2

d=√(xa-xb)^2 +(ya-yb)^2

d=√(2u0cos theta t-l)^2+h^2

dmin when (2ucos theta t-l)=minimum

dminimum=h

3) you can try it by i told these two

ALL THE BEST .

Answer 2

Explanation:

1) VAx=u0cos theta

VBx=-u0cos theta

VABx=u0cos theta-gt

VBY= u0 sin theta -gt

i.e VABY=0

SO WRT B A IS moving in the straight line as it has only constant velocity VABX=2U0COS THETA along x - axis

2) at any time t

xA=u0cos theta

xb=l-u cos theta-t

ya=usin theta t-1/2 gt^2

yb= -d(usin thetat-1/2 gt^2

d=√(xa-xb)^2 +(ya-yb)^2

d=√(2u0cos theta t-l)^2+h^2

dmin when (2ucos theta t-l)=minimum

dminimum=h

3) you can try it by i told these two