Question

A projectile cover double range as compare to its

maximum height attained. The angle of projection

from horizontal is–​

Answer 1

Given:

Projectile range is double in magnitude as compared to the maximum height reached by that object

To find:

Angle of projection.

Calculation:

Let initial velocity be u , gravity be g and angle of projection be $\theta$

$range = 2 \times (max \: height)$

$= > \dfrac{ {u}^{2} \sin(2 \theta) }{g} = 2 \times \bigg \{ \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g} \bigg \}$

$= > \dfrac{ {u}^{2} \sin(2 \theta) }{g} = \cancel2 \times \bigg \{ \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{ \cancel{2}g} \bigg \}$

$= > \dfrac{ \cancel{ {u}^{2}} \sin(2 \theta) }{ \cancel g} = \dfrac{ \cancel{{u}^{2} } { \sin}^{2}( \theta) }{ \cancel g}$

$= > \sin(2 \theta) = { \sin}^{2} ( \theta)$

$= > 2 \sin( \theta) \cos( \theta) = { \sin}^{2} ( \theta)$

$= > \dfrac{ \sin( \theta) }{ \cos( \theta) } = 2$

$= > \tan( \theta) = 2$

$= > \theta = { \tan}^{ - 1} (2)$

So , angle of Projection is

$\boxed{ \red{ \bold{\theta = { \tan}^{ - 1} (2)}}}$

Answer 2

$\large \underline{\underline{\mathfrak{Answer : }}}$

• Angle of projection is $\sf{\tan^{-1} (2)}$

$\underline{\underline{\mathfrak{Step-By-Step -Explanation \: :}}}$

GiveN :

• A projectile covers double range as compared to its maximum hight.

To FinD :

• Angle of Projection

SolutioN :

$\longrightarrow \sf{Range \: = \: 2(Max. \: Height)} \\ \\ \longrightarrow \sf{\dfrac{u^2 \sin 2 \theta}{g} \: = \: \cancel{2} \bigg( \dfrac{u^2 \sin ^2 \theta}{\cancel{2}g} \bigg) } \\ \\ \longrightarrow \sf{\dfrac{\cancel{u^2} \sin 2 \theta}{\cancel{g}} \: = \: \dfrac{\cancel{u^2} \sin ^2 \theta }{\cancel{g}}} \\ \\ \longrightarrow \sf{\sin 2 \theta \: = \: \sin ^2 \theta} \\ \\ \longrightarrow \sf{2 \cancel{\sin \theta} \cos \theta \: = \: \sin^{\cancel{2}} \theta} \\ \\ \longrightarrow \sf{2 \cos \theta \: = \: \sin \theta } \\ \\ \longrightarrow \sf{2 \: = \: \dfrac{\sin \theta}{\cos \theta}} \\ \\ \longrightarrow \sf{\tan \theta \: = \: 2} \\ \\ \Large \underline{\boxed{\sf{\theta \: = \: \tan^{-1} (2)}}}$

_________________________________

$\boxed{\begin{minipage}{5cm} {\large{\underline{\underline{\textsf{\textbf{Formulae \: Used \: :}}}}}} \\ \\ \bullet \: \sf{R \: = \: \dfrac{u^2 \sin 2 \theta}{g}} \\ \\ \bullet \: \sf{H_{max} \: = \: \dfrac{u^2 \sin ^2 \theta}{2g}} \\ \\ \bullet \: \sf{\sin 2 \theta \: = \: 2 \sin \theta \cos \theta} \\ \\ \bullet \: \sf{\tan \theta \: = \: \dfrac{\sin \theta}{\cos \theta}} \end{minipage}}$