Answer:
Let u be
the initial velocity of projection of body and v be the velocity of the same
body while passing downwards through point of projection. The displacement of
body s = 0. Using the relation\[{{\text{v}}^{\text{2}}}=\text{
}{{\text{u}}^{\text{2}}}+\text{ 2 as}\], and\[\text{u}=\text{u},\text{
v}=?\text{ };\text{a}=-\text{g},\text{ s}=0\], we have \[{{\text{v}}^{\text{2}}}=\text{
}{{\text{u}}^{\text{2}}}+\text{ 2}\left( -\text{g} \right)\text{ }\times \text{
}0\text{ }={{\text{u}}^{\text{2}}}\text{orv}=\text{u}\]
It means the final speed is independent of mass of the
body. Hence both the bodies will acquire the same speed while passing through
point of projection.
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