Question

A particle is projected such that the horizontal range and vertical height are the same. Then the angle of projection is :
I need detailed explaination..​

Answer 1

Answer:

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)