In the given figure for a projectile,find the | vedclass

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(B) The equation of the trajectory of a projectile is given by:$y = x 	an 	heta \left(1 - \frac{x}{R}ight)$where $R$ is the horizontal range.From

Vedclass · Accepted Answer

$y=\left[\frac{x_1 x_2}{x_1+x_2}ight] 	an 	heta$

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(B) The equation of the trajectory of a projectile is given by:$y = x 	an 	heta \left(1 - \frac{x}{R}ight)$where $R$ is the horizontal range.From the figure,the total range $R = x_1 + x_2$.At point $P$,the horizontal distance is $x = x_1$.Substituting these values into the trajectory equation:$y = x_1 	an 	heta \left(1 - \frac{x_1}{x_1 + x_2}ight)$$y = x_1 	an 	heta \left(\frac{x_1 + x_2 - x_1}{x_1 + x_2}ight)$$y = x_1 	an 	heta \left(\frac{x_2}{x_1 + x_2}ight)$$y = \left[\frac{x_1 x_2}{x_1 + x_2}ight] 	an 	heta$Thus,the correct option is $B$.

In the given figure for a projectile,find the relation for $y$ in terms of $x_1, x_2$ and $\theta$.

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