Question

For a projectile motion ground to ground maximum height is equal to horizontal range . Find the angle of projection.

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

For a projectile motion ground to ground maximum height is equal to horizontal range . Find the angle of projection?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Let the angle at the Projectile is projected be θ.
• And, The range & Height of Projectile be "R" and "H" respectively.

$\huge{\bold{\underline{Explanation:-}}}$

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As Mentioned in the above Question,

$\large{\boxed{\tt H_{max} = R}}$

Substituting the Respective Formulas,

$\large{\tt \leadsto \dfrac{u^2 sin^2 \theta}{2g} = \dfrac{u^2 sin 2\theta}{g}}$

$\large{\tt \leadsto \dfrac{u^2 sin^2 \theta}{2\cancel{g}} = \dfrac{u^2 sin 2\theta}{\cancel{g}}}$

$\large{\tt \leadsto \dfrac{u^2 sin^2 \theta}{2} = u^2 sin 2\theta}$

As we Know,

Sin2θ = 2sinθcosθ,

Substituting it,

$\large{\tt \leadsto \dfrac{u^2 sin^2 \theta}{2} = u^2 \times 2sin \theta cos \theta}$

$\large{\tt \leadsto \dfrac{\cancel{u^2} sin^2 \theta}{2} = \cancel{u^2} \times 2sin \theta cos \theta}$

$\large{\tt \leadsto \dfrac{sin^2 \theta}{2} = 2sin \theta cos \theta}$

$\large{\tt \leadsto \dfrac{sin\theta \times sin\theta}{2} = 2sin \theta cos \theta}$

$\large{\tt \leadsto \dfrac{sin\theta \times \cancel{sin\theta}}{2} = 2\cancel{sin \theta} cos \theta}$

$\large{\tt \leadsto \dfrac{sin\theta }{2} = 2 cos \theta}$

Rearranging,

$\large{\tt \leadsto \dfrac{sin \theta}{cos \theta} = 2 \times 2}$

As we know,

Sinθ/Cosθ = Tanθ, Substituting it,

$\large{\tt \leadsto Tan\theta = 4}$

$\huge{\boxed{\boxed{\tt \theta = Tan^{-1} (4) \degree}}}$

So,the Angle of Projection is θ = Tan⁻¹(4)°.

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Answer 2

$\Huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

We are Given,

• In an angular projection Maximum height and horizontal range are equal.

And we have to find Angle of projection.

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Formula for Maximum Height is :-

$\LARGE{\boxed{\boxed{\green{\bf{h \: = \: \frac{u^2 \: Sin^2 \theta}{2g}}}}}}$

And for Horizontal Range is :-

$\LARGE{\boxed{\boxed{\orange{\bf{R \: = \: \frac{u^2 \: Sin2 \theta}{g}}}}}}$

A.T.Q,

Maximum height and Horizontal Range are equal.

$\Large \leadsto {\sf{\frac{\cancel{u^2}Sin^2 \theta}{2 \cancel{g}} \: = \: \frac{\cancel{u^2} Sin 2 \theta }{\cancel{g}}}}$

$\Large \leadsto {\sf{\frac{Sin^2 \theta}{2} \: = \: \frac{Sin 2 \theta}{1}}}$

$\Large \leadsto {\sf{\frac{Sin \theta \: \times \: \cancel{Sin \theta}}{2} \: = \: 2 \cancel{Sin \theta} Cos \theta }}$

$\Large \leadsto {\sf{\frac{Sin \theta }{2} \: = \: 2 Cos \theta}}$

$\Large \leadsto {\sf{\frac{Sin \theta }{4} \: = \: Cos \theta}}$

$\Large \leadsto {\sf{\frac{Sin \theta }{Cos \theta} \: = \: 4}}$

$\Large \leadsto {\sf{Tan \theta \: = \: 4}}$

$\Large \leadsto {\sf{ \theta \: = \: Tan^{-1}( 4)}}$

$\LARGE{\boxed{\boxed{\red{\sf{ \theta \: = \: 75.963^{\circ}}}}}}$