Respuesta :

jaseelk92

Answer:

The value of [tex]\theta=63.43^0[/tex]

Explanation:

Range of projectile [tex]\mathrm{R}=\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}[/tex]

Height of projectile [tex]\mathrm{h}=\frac{\mathrm{u}^{2} \cdot \sin ^{2} \theta}{2 \cdot \mathrm{g}}[/tex]

Here we have

[tex]R=2h\\\\\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}=2\frac{\mathrm{u}^{2} \cdot \sin ^{2} \theta}{2 \cdot \mathrm{g}}\\\\\sin 2 \theta=\sin ^{2} \theta\\\\2\sin \theta\cos \theta=\sin ^{2} \theta\\\\\tan \theta=2\\\\\theta=63.43^0[/tex]

Range be R and height be h

[tex]\boxed{\sf R=\dfrac{u^2sin2\theta}{g}}[/tex]

[tex]\boxed{\sf h=\dfrac{u^2sin^2\theta}{2g}}[/tex]

  • u=initial velocity
  • theta is angle of projection.
  • g=acceleration due to gravity

ATQ

[tex]\\ \sf\longmapsto R=2h[/tex]

[tex]\\ \sf\longmapsto \dfrac{u^2sin2\theta}{g}=\dfrac{2u^2sin^2\theta}{2g}[/tex]

  • Cancelling required ones

[tex]\\ \sf\longmapsto sin^2\theta=sin2\theta[/tex]

  • sin2O=2sinOcosO

[tex]\\ \sf\longmapsto sin^2\theta=2sin\theta cos\theta [/tex]

[tex]\\ \sf\longmapsto \dfrac{sin^2\theta}{sin\theta cos\theta=2[/tex]

[tex]\\ \sf\longmapsto \dfrac{sin\theta}{cos\theta}=2[/tex]

[tex]\\ \sf\longmapsto tan\theta=2[/tex]

[tex]\\ \sf\longmapsto \theta=tan^{-1}(2)[/tex]

[tex]\\ \sf\longmapsto \theta=63.4°[/tex]

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