Question

a particle is projected with a velocity u making an angle X with the horizontal .At any instant it's velocity v is at right angles to its initial velocity u then v - 1 u cosx 2 u tanx 3 ucotx 4 usecx give explanation

Answer 1

initial velocity(u)=ucosx i^+usinx j^
V=ucosx i^+(usinx-gt) j^=========(1)

a/c to question ,
u.V=0 {dot product always zero when two vectors are perpendicular }
u^2cos^2x+u^2sin^2x-usinxgt=0
u^2{sin^2x+cos^2x}-usinx.gt=0
u^2-usinx.gt=0
t=u/gsinx
put this equation (1)

V=ucosx i^+(usinx-gu/gsinx) j^
=ucosx i^+(usinx-u/sinx) j^
=ucosx i^-ucos^2x/sinx j^
now magnitude of this vector
|V|=√{u^2cos^2x+u^2cos^4x/sin^2x}
=u√{cos^2(sin^2x+cos^2x}/sin^2x}
=ucotx