In the given figure for a projectile

Question

In the given figure for a projectile (A) y=[(x1 x2/x1-x2)] tan θ (B) y=[(x1 x2/x1+x2)] tan θ (C) y=[(2 x1 x2/x1+x2)] cos θ (D) y=[(2 x1 x2/x1+x2)] tan θ

Tardigrade Admin · Accepted Answer

The equation of trajectory for point 'P' can be written as: y=x tan θ(1-(x/R))=x1 tan θ(1-(x1/x1+x2)) =x1 tan θ((x1+x2-x1/x1+x2)) y=(x1 x2/x1+x2) tan θ

Q. In the given figure for a projectile

$y = [\frac{x _{1} x _{2}}{x _{1} - x _{2}}] tan θ$

$y = [\frac{x _{1} x _{2}}{x _{1} + x _{2}}] tan θ$

$y = [\frac{2 x _{1} x _{2}}{x _{1} + x _{2}}] cos θ$

$y = [\frac{2 x _{1} x _{2}}{x _{1} + x _{2}}] tan θ$

The equation of trajectory for point 'P' can be written as :
$y = x tan θ (1 - \frac{x}{R}) = x_{1} tan θ (1 - \frac{x _{1}}{x _{1} + x _{2}})$
$= x_{1} tan θ (\frac{x _{1} + x _{2} - x _{1}}{x _{1} + x _{2}})$
$y = \frac{x _{1} x _{2}}{x _{1} + x _{2}} tan θ$

Q. In the given figure for a projectile

y=[x1​−x2​x1​x2​​]tanθ

y=[x1​+x2​x1​x2​​]tanθ

y=[x1​+x2​2x1​x2​​]cosθ

y=[x1​+x2​2x1​x2​​]tanθ

The equation of trajectory for point 'P' can be written as : y=xtanθ(1−Rx​)=x1​tanθ(1−x1​+x2​x1​​) =x1​tanθ(x1​+x2​x1​+x2​−x1​​) y=x1​+x2​x1​x2​​tanθ

$y = [\frac{x _{1} x _{2}}{x _{1} - x _{2}}] tan θ$

$y = [\frac{x _{1} x _{2}}{x _{1} + x _{2}}] tan θ$

$y = [\frac{2 x _{1} x _{2}}{x _{1} + x _{2}}] cos θ$

$y = [\frac{2 x _{1} x _{2}}{x _{1} + x _{2}}] tan θ$

The equation of trajectory for point 'P' can be written as :
$y = x tan θ (1 - \frac{x}{R}) = x_{1} tan θ (1 - \frac{x _{1}}{x _{1} + x _{2}})$
$= x_{1} tan θ (\frac{x _{1} + x _{2} - x _{1}}{x _{1} + x _{2}})$
$y = \frac{x _{1} x _{2}}{x _{1} + x _{2}} tan θ$