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STatiana [176]
3 years ago
7

Can someone explain it please ?​

Physics
2 answers:
tino4ka555 [31]3 years ago
7 0

Range be R and height be h

\boxed{\sf R=\dfrac{u^2sin2\theta}{g}}

\boxed{\sf h=\dfrac{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 \dfrac{u^2sin2\theta}{g}=\dfrac{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 \dfrac{sin^2\theta}{sin\theta cos\theta=2

\\ \sf\longmapsto \dfrac{sin\theta}{cos\theta}=2

\\ \sf\longmapsto tan\theta=2

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

\\ \sf\longmapsto \theta=63.4°

liq [111]3 years ago
4 0

Answer:

The value of \theta=63.43^0

Explanation:

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

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

Here we have

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

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