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Intro to Relative Projectile

A particle is...

Question

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:

A

θ upwards

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B

θ downwards

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C

Zero

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D

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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