Question

Let angle of projection be $\theta$ , angle of elevation to a point on the trajectory (from point of projection) be $\alpha$ and angle of elevation to the same point (from the point where projectile meets the ground) be $\beta$.

PROVE:
$\boxed{ \tan( \theta) = \tan( \alpha ) + \tan( \beta ) }$


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Answer 1

$\large\underline{\red{\sf \pink{\bigstar} Correct\; Question}}$

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• ➩ Let angle of projection be $\theta$ , angle of elevation to a point on the trajectory (from point of projection) be $\alpha$ and angle of elevation to the same point (from the point where projectile meets the ground) be $\beta$.

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• ➩ $\pink{\sf{\star\;To \;Prove}}$

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• ➩ $\boxed{ \tan( \theta) = \tan( \alpha ) + \tan( \beta ) }$

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$\large\underline{\blue{\sf \purple{\bigstar} Solution:-}}$

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• ➩ Let p be the point on trajectory, that subtends angle α, β at the line point of projection and landing

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• ➩ let y be the height of p from horizontal

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$\red{\sf{\star\;Now}}$

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• ➩ tan α. + tan β = $\frac{y}{x}$ +$\frac{y}{R-x}$

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• ➩ tan α +tan β = $\frac{yR}{x(R-x)}$

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$\pink{\sf{\star\;Now\; equation \;of \;trajectory \;is }}$

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• ➩ y = xtan θ ( 1- xyR)

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• ➩ tan θ = $\frac{yR}{x ( R - x)}$

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$\orange{\sf{\star\;Hence}}$ $\boxed{ \tan( \theta) = \tan( \alpha ) + \tan( \beta ) }$

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$\green{\sf{\star\;Hence\; Proved !}}$

Answer 2

$\bold{Answer\;:-}$

• Given :-

• To prove : tan α + tan β = tan θ

• Solution :-

tan α = y/x

tan β = y/(R-x)

Therefore, tan α + tan β = y/x + y/R-x

• Equation of trajectory ( in the form of range ) :-

y = x*tan θ [1-(x/R)]

y/x = tan θ [(R-x)/R]

y*R/x(R-x) = tan θ - (i)

Now, tan α + tan β = y/x + y/(R-x)

tan α + tan β = y [ (R-x+x)/x(R-x)]

tan α + tan β = y*R/x(R-x) - (ii)

Now, from equation (i)

tan α + tan β = tan θ

Hence proved!

Extra :-

• Eqn. of trajectory : [ Refer to the attachment! ]

• Projectile Motion :-

• A particle which is thrown into the space and which moves under the effect of gravity only is called a projectile.

• Projectile motion is a combination of two motions : Horizontal motion and Vertical motion.

• The acceleration due to gravity causes the ball to slow down as it goes upwards.

• There is no force in the horizontal direction so there is no acceleration.

• Trajectory of a projectile is parabolic only when the acceleration of the projectile is constant and the direction of acceleration is different from the direction of projectile's initial velocity.

• Under the same acceleration due to gravity, the trajectory of an object can be a straight line or a parabola depending on the initial conditions.

$\bold{Hope\;it \; helps\;!}$