Question

find the angle of projection of a body for which its horizontal range is equal to the maximum height reached by it

Answer 1

GIVEN :-

• Horizontal range is equal to maximum height

TO FIND :-

• Angle of projection of the body

SOLUTION :-

Range of a ptojectile is given by ,

$\large \underline {\bold {\boxed {\bigstar{\red {\sf{ \: r = \frac{ {u}^{2} \sin(2 \theta) }{g}}}}}}}$

Maximum height of a projectile is given by,

$\large {\underline {\bold {\boxed {\bigstar {\red {\sf{ \: h = \frac{ {u}^{2} \sin {}^{2} ( \theta) }{2g} }}}}}}}$

Where ,

• u is initial velocity
• θ is angle of projection
• g is acceleration due to gravity

According to the question ,

$\implies \sf \dfrac{ {u}^{2} \sin(2 \theta) }{g} = \dfrac{ {u}^{2} \sin {}^{2} ( \theta) }{2g} \\ \\ \implies \sf \: \frac{ \cancel{ {u}^{2}} \sin(2 \theta) }{ \cancel{g}} = \frac{ \cancel{ {u}^{2} } \sin {}^{2} ( \theta) }{2 \cancel{g}} \\ \\ \implies \sf \: \sin( 2\theta) = \frac{ \sin {}^{2} ( \theta) }{2} \\ \\ \implies \sf \: 2 \sin(2 \theta) = \sin {}^{2} ( \theta)$

$\large {\underline {\bold {\boxed {\bigstar {\red {\sf{ \: \sin(2 \theta) = 2 \sin( \theta) \cos( \theta) }}}}}}}$

$\implies \sf \: 2[2 \sin( \theta) \cos( \theta) ] = \sin { }^{2} ( \theta) \\ \\ \implies \sf \: 4 \cancel{\ sin( \theta)} \cos( \theta) = \cancel{ \sin {}^{2} ( \theta) }\\ \\ \implies \sf \: 4 \cos( \theta) = \sin( \theta) \\ \\ \implies \: 4 = \frac{ \sin( \theta) }{ \cos( \theta) }$

$\large {\underline {\bold {\boxed {\bigstar {\red {\sf{ \: \tan(\theta) = \dfrac{\sin( \theta)}{ \cos( \theta)} }}}}}}}$

$\implies \sf \: 4 = \tan( \theta) \\ \\ \implies \sf \theta = \tan {}^{ - 1} (4)$

∴ The angle of projection of the body is Tan⁻¹ (4)