Question

Find the angle of projection at which horizontal range and maximum height are equal.​

Answer 1

Answer:

Given

horizontal range = maximum height

$range = \frac{ {u}^{2} \sin(2 \alpha ) }{g} \\ height = \frac{ {u}^{2} { \sin( \alpha ) }^{2} }{2g} \\ but \\ \frac{ {u}^{2} \sin(2 \alpha ) }{g} = \frac{ {u}^{2} { \sin( \alpha ) }^{2} }{2g} \\ 2 \sin( \alpha ) \cos( \alpha ) = \frac{ { \sin( \alpha ) }^{2} }{2} \\ 2 \cos( \alpha ) = \frac{ \sin( \alpha ) }{2} \\ \tan( \alpha ) = 4 \\ \alpha = { \tan }^{ - 1} 4$