Question

Prove that for two complimentary angle of projection of projectile thrown with
same velocity, horizontal range are equal.
....Plz answer sum1
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Answer 1

Question

Prove that for two complimentary angle of projection of projectile thrown with same velocity, horizontal range are equal.

Solution

The projectile have same initial velocities and difference projection angles

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$\boxed{ \boxed{ \sf \: R = \dfrac{ {u}^{2} \sin(2 \alpha ) }{g}}}$

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The angles are complimentary,i.e.,they add up to 90°.

Thus,one angle would be $\alpha$ and the other angle would be $90 - \alpha$

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When the angle is is $\alpha$,

$\sf \: R_1 = \dfrac{ {u}^{2} \sin(2 \alpha ) }{g} - - - - - - - (1)$

When the angle is $90 - \alpha$,

$\sf \: R_2 = \dfrac{ {u}^{2} \sin(2(90 - \alpha )) }{g} \\ \\ \longrightarrow \: \sf \: R_2 = \dfrac{ {u}^{2} \sin(180 - 2 \alpha ) }{g} \\ \\ \longrightarrow \: \sf \: R_2 = \dfrac{ {u}^{2} \sin(2 \alpha ) }{g} - - - - - - - (2)$

From equations (1) and (2),we write :

Projectile projected with two different complementary angles have same horizontal ranges

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Answer 2

Answer:

Given:

2 Projectiles are thrown with same Velocity and at complimentary angles.

To prove:

Their horizontal Projectiles are same.

Formulas used:

Let θ be angle of Projection, v is Velocity of Projection , g is gravity , then

$\: \: \: \: \: \: \: \: \boxed{ \sf{ \red{ \large{Range = \dfrac{ {u}^{2} \sin(2 \theta) }{g}}}}}$

Calculation:

Let 1st angle of Projection be θ , then

$\therefore \: R_{1} = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

Now the 2nd Projectile is thrown at complimentary angle ; i.e tge angle of Projection is (90° - θ)

$\therefore \: R_{2} = \dfrac{ {u}^{2} \sin \{2(90 \degree - \theta) \} }{g}$

$\implies R_{2} = \dfrac{ {u}^{2} \sin (180 \degree - 2\theta) }{g}$

$\implies R_{2} = \dfrac{ {u}^{2} \sin ( 2\theta) }{g}$

$\implies R_{2} = R_{1}$

So the ranges are equal , provided the Velocity of Projection is same in both the cases.

$\boxed{ \sf{ \large{ \red{hence \: proved}}}}$