In the given figure for a projectile - Vidyadip

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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