A particle is projected from point A on the ground at an angle $θ$ with horizontal. Another particle is projected from point B simultaneously from height 4H above point A with same velocity. A and B are in the same vertical plane, where H is the maximum height attained by the particle A. Both the particles collide at the same time on the ground. Find the angle at which particle B is projected with horizontal:

θ upwards

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θ downwards

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Zero

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2θ upwards

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Solution

The correct option is C
Zero

h=4H= $\frac{4 u^{2} s i n^{2} θ}{2 g}$

= $\frac{1}{2} . g . \frac{4 u^{2} s i n^{2} θ}{g^{2}}$

= $\frac{1}{2} . g . {(\frac{2 u s i n θ}{g})}^{2} = \frac{g}{2} t_{A}^{2}$
$t_{A} = t i m e o f f l i g h t o f A$
$t_{A} = \sqrt{\frac{2 h}{g}}$

Since particles collide,

$t_{A} = t_{B}$

Then, $t_{A} = t_{B} = \sqrt{\frac{2 h}{g}}$

This can happen only if the particle B is projected horizontally.

Or $θ = 0$

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The correct option is C Zero h=4H=4u2sin2θ2g =12.g.4u2sin2θg2 =12.g.(2usinθg)2=g2t2A tA=timeofflightofA tA=√2hg Since particles collide, tA=tB Then, tA=tB=√2hg This can happen only if the particle B is projected horizontally. Or θ=0