Question

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Answer 1

$\huge\bigstar\underline\pink{Answer}$

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★ Hope you know derivation of equation of trajectory

$\large\blue{equation\:of\: trajectory}$

Y = x1 tan ∅ [ 1- x1/R]. .......(1)

→ R = x1 + x2

putting the value of R in 1st equation

★ Y = x1 tan ∅ [ 1- x1/(x1+x2)]

y = x1 tan ∅ [ x1 + x2 - x1/ x1 + x2]

y = x1 tan ∅ (x2/ x1 - x2 )

y = (x1x2/x1+x2)tan∅

$\boxed{option\:2\:is\: correct}$

Answer 2

Answer:

$\displaystyle{y=\left[\dfrac{x_1x_2}{x_1+x_2}\right]\tan\theta}$

Option 2.  is correct .

Explanation:

We have :

Range ( R ) = u² sin 2 θ / g        [ sin 2 θ = 2 sin θ cos θ  ]

R = u² 2 sin θ cos θ / g

Taking reciprocal :

1 / R = g / u² 2 sin θ cos θ   .... ( i )

We have equation of Trajectory :

$\displaystyle{y=x \ \tan\theta-\left[\dfrac{g}{2 \ u^2\cos^2\theta}\right]x^2}$

Multiply ans divide by sin θ in x² term

$\displaystyle{y=x \ \tan\theta-\left[\dfrac{g}{2 \ u^2\cos^2\theta\times\dfrac{\sin\theta}{\sin\theta} }\right]x^2}\\\\\\\displaystyle{y=x \ \tan\theta-\left[\dfrac{g}{2 \ u^2\sin\theta\cos\theta\times\dfrac{\cos\theta}{\sin\theta} }\right]x^2}$

From ( i )  we have :

1 / R = g / u² 2 sin θ cos θ  

$\displaystyle{y=x \ \tan\theta-\left[\dfrac{\tan\theta}{\text{R}}\right]x^2}\\\\\\\displaystyle{y=\tan\theta\left[x-\dfrac{x^2}{\text{R}} \right]}$

Here we have  R  =  x₁  +  x₂   &  x =  x₁

$\displaystyle{y=\tan\theta\left[x_1-\dfrac{x_1^2}{x_1+x_2} \right]}\\\\\\\displaystyle{y=\tan\theta\left[x_1\left(1-\dfrac{x_1}{x_1+x_2}\right) \right]}\\\\\\\displaystyle{y=\tan\theta\left[x_1\left(\dfrac{x_1+x_2-x_1}{x_1+x_2}\right) \right]}\\\\\\\displaystyle{y=\tan\theta\left[x_1\left(\dfrac{x_2}{x_1+x_2}\right) \right]}$

$\displaystyle{y=\tan\theta\left[\dfrac{x_1x_2}{x_1+x_2}\right]}$

Hence we get answer .