Question

can someone explain it please ?​

Answer 1

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

$R=2h\\\\\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}=2\frac{\mathrm{u}^{2} \cdot \sin ^{2} \theta}{2 \cdot \mathrm{g}}\\\\\sin 2 \theta=\sin ^{2} \theta\\\\2\sin \theta\cos \theta=\sin ^{2} \theta\\\\\tan \theta=2\\\\\theta=63.43^0$

Answer 2

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

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

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

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

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

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

$\\ \sf\longmapsto \theta=63.4°$