Question

a particle is projected such that the horizontal range and vertical height are equal. then the angle of projection is

Answer 1

Answer:

$tan \: theta$ = 4

$theta$ =tan ^-1 (4)

Answer 2

Hey mates your answer is here

according to the question,

horizontal range = vertical height

we know,

horizontal range = \frac{u^2sin2{\theta}}{g}

vertical height = \frac{u^2sin^2{\theta}}{2g}

so,

\frac{u^2sin2{\theta}}{g}=\frac{u^2sin^2{\theta}}{2g} \\ 2sin{\theta}.cos{\theta}=\frac{sin^2{\theta}}{2} \\ tan{\theta} = 4 \\ {\theta}=tan^{-1}(4)

May be it's helpful for you .

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