Question

que
0. Prove that ranges are same for two angle of projections which are complementary to
each other​

Answer 1

$\sf\red{Answer - }$

Range is the horizontal distancd covered by object during projectile motion -

$\implies$$\bf\purple{ R = u_x \times t}$

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$\bf\pink{ u_x = ucos\theta }$

$\bf \pink{t = \frac{2usin\theta}{g}}$

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$\implies$$\bf R = ucos\theta \times \frac{2usin\theta}{g}$

$\implies$$\bf R = \frac{u^2(2sin\theta cos\theta)}{g}$

$\implies$$\bf\red{ R = \frac{u^2sin2\theta}{g}}$

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$\bf Replacing \theta = 90 - \theta$ (complementary angle for $\theta$)

$\implies$$\bf R' = \frac{u^2sin2(90 - \theta) }{g}$

$\implies$$\bf R' = \frac{u^2sin(180-2\theta)}{g}$

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$\implies$$\bf sin(180-2\theta) = sin 2\theta$ ( because 180° is non changing line )

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$\implies$$\bf R' = \frac{u^2sin2\theta}{g}$

$\bf\purple{ R' = R}$

Hence we get the same value for the angle amd its complementary for range.

Hence Proved.

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Additional information -

Maximum range -

$\implies$$\bf R' = \frac{u^2sin2\theta}{g}$

For max. range

$\implies$$\bf sin2\theta = 1$

$\implies$$\bf 2\theta = 90\degree$

$\implies$$\bf \theta = 45\degree$

$\implies$$\bf \red{R_{max} = \frac{u^2}{g}}$

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Maximum height -

Using 3rd equation of motion in y direction

$\bf v^2 = u^2 + 2as$

$\implies$$\bf v_y^2 = u_y^2 + 2gh$

For maximum height

$\longrightarrow$$\bf v_y = 0$

$\longrightarrow$$\bf h = H$

$\implies$$\bf 0 = (-usin\theta)^2 + 2g(-H)$

$\implies$$\bf\red{H = \frac{u^2sin^2\theta}{g}}$

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