MCQ
In the given figure for a projectile
  • A
    $y=\left[\frac{x_1 x_2}{x_1-x_2}\right] \tan \theta$
  • $y=\left[\frac{x_1 x_2}{x_1+x_2}\right] \tan \theta$
  • C
    $y=\left[\frac{2 x_1 x_2}{x_1+x_2}\right] \cos \theta$
  • D
    $y=\left[\frac{2 x_1 x_2}{x_1+x_2}\right] \tan \theta$

Answer

Correct option: B.
$y=\left[\frac{x_1 x_2}{x_1+x_2}\right] \tan \theta$
b
(b)

The equation of trajectory for point ' $P$ 'can be written as :

$y=x \tan \theta\left(1-\frac{x}{R}\right)=x_1 \tan \theta\left(1-\frac{x_1}{x_1+x_2}\right)=x_1 \tan \theta\left(\frac{x_1+x_2-x_1}{x_1+x_2}\right)$

$y=\frac{x_1 x_2}{x_1+x_2} \tan \theta$

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