Question

a particle is projected such that the horizontal range and vertical height are the same .Then the angle of projection is?

Answer 1

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

Answer 2

$\large\large{\sf{\purple{\boxed{\boxed{\pink{ANSWER}}}}}}$

→using the realtion,

$If \: R \: = \: n \: H \\ \\ then \: tan\theta \: = \: \frac{4}{n}$

here in the ques ,

• n = 1

$tan\theta \: = \: 4 \\ \\ \theta \: = {tan}^{ - 1} 4$