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Question

(B) Let $\vec{u}_{1}$ and $\vec{u}_{2}$ be the initial velocities of the two particles and $	heta_{1}$ and $	heta_{2}$ be their angles of projection

Vedclass · Accepted Answer

a straight line making a constant angle with the horizontal

Vedclass · Answer

(B) Let $\vec{u}_{1}$ and $\vec{u}_{2}$ be the initial velocities of the two particles and $	heta_{1}$ and $	heta_{2}$ be their angles of projection with the horizontal.The velocities of the two particles at any time $t$ are given by:$\vec{v}_{1} = (u \cos 	heta_{1}) \hat{i} + (u \sin 	heta_{1} - gt) \hat{j}$$\vec{v}_{2} = (u \cos 	heta_{2}) \hat{i} + (u \sin 	heta_{2} - gt) \hat{j}$The relative velocity of one particle with respect to the other is:$\vec{v}_{12} = \vec{v}_{1} - \vec{v}_{2} = u(\cos 	heta_{1} - \cos 	heta_{2}) \hat{i} + u(\sin 	heta_{1} - \sin 	heta_{2}) \hat{j}$Since the relative velocity $\vec{v}_{12}$ is independent of time $t$,the relative acceleration is zero $(\vec{a}_{12} = \vec{a}_{1} - \vec{a}_{2} = -g\hat{j} - (-g\hat{j}) = 0)$.Because the relative acceleration is zero and the relative initial position is zero (both start from the same point),the relative motion is a straight line passing through the origin. Since the relative velocity vector is constant,the path is a straight line making a constant angle with the horizontal.

$Column-I$	$Column-II$
$(A)$ Angle of projection	$(p)$ $20 \, m$
$(B)$ Angle of velocity with horizontal after $4 \, s$	$(q)$ $80 \, m$
$(C)$ Maximum height	$(r)$ $45^{\circ}$
$(D)$ Horizontal range	$(s)$ $\tan^{-1}(1/2)$

Two particles are projected simultaneously in the same vertical plane,from the same point on the ground,with the same speeds but at different angles $( < 90^o )$ to the horizontal. The path followed by one,as seen by the other,is

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