Question

an object is projected with velocity v. if the range of this object is double the maximum height H, then it's range is​

Answer 1

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H = $\frac{R}{4}Tanθ$

Given: R = 2H

⇒ $H=\frac{H}{2}Tanθ$

⇒ $Tanθ=2$

If Tanθ = 2 then

Sinθ =  $\frac{2}{√5}$

Cosθ = $\frac{1}{\sqrt{5} }$

$R=\frac{2u² sinθ cosθ }{g}$

=  $\frac{(2u² × 2/5 ) }{g}$

=  $\frac{4u^2}{5 g}$

Range of projectile is $\frac{4u^2}{5 g}$

Hope it Helps !!!