1 Answer

MrKsn · Answer 1 · 2022-01-21T02:35:07+0000

Range be R and height be h

$\boxed{\sf R=(u^2sin2\theta)/(g)}$

$\boxed{\sf h=(u^2sin^2\theta)/(2g)}$

u=initial velocity

theta is angle of projection.

g=acceleration due to gravity

ATQ

$\\ \sf\longmapsto R=2h$

$\\ \sf\longmapsto (u^2sin2\theta)/(g)=(2u^2sin^2\theta)/(2g)$

Cancelling required ones

$\\ \sf\longmapsto sin^2\theta=sin2\theta$

sin2O=2sinOcosO

$\\ \sf\longmapsto sin^2\theta=2sin\theta cos\theta$

$\\ \sf\longmapsto \frac{sin^2\theta}{sin\theta cos\theta=2$

$\\ \sf\longmapsto (sin\theta)/(cos\theta)=2$

$\\ \sf\longmapsto tan\theta=2$

$\\ \sf\longmapsto \theta=tan^(-1)(2)$

$\\ \sf\longmapsto \theta=63.4°$

Done

Option B is correct

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