Question

For same speed of projection range are equal for angle theta height is h1 and for other is h2 proove that range = root 16 h1 h2

Answer 1

$\Large{\sf{\green{Given}}}$

• Speed for both cases are equal

• (h₁) height attained by θ projection

• (h₂) height attained by another projection

$\Large{\sf{\green{To\: prove}}}$

• $\sf{Range\: (R) = \: \sqrt{16\:h_{1}h_{2}} = \: \: 4\sqrt{h_{1}h{2}}}$

$\Large{\sf{\green{Solution}}}$

Finding Range (R)

$\large{\sf{R = \frac{u^{2}sin2\theta}{g}}\: \: \blue{\longrightarrow\: (1)}}$

Let h₁ and h₂ be the maximum height attained by θ and (90 - θ) projection respectively (Reason is given below)

$\large{\sf{h_{1} = \frac{u^{2}sin^{2}\theta}{2g}\: \: \red{\longrightarrow\:(2)}}}$

$\large{\sf{h_{2} = \frac{u^{2}sin^{2}(90 - \theta)}{2g}\: \: \red{\longrightarrow\:(3)}}}$

Multiplying (2) and (3) :

$\sf{\implies\: h_{1} . h_{2} = \frac{u^{2}sin^{2}\theta}{2g} × \frac{u^{2}sin^{2}(90 - \theta)}{2g}}$

$\sf{\implies\: h_{1} . h_{2} = \frac{u^{2}sin^{2}\theta}{2g} × \frac{u^{2}cos^{2} \theta}{2g}}$

$\sf{\implies\: h_{1} . h_{2} = \frac{u^{4}sin^{2}\theta cos^{2} \theta}{4g^{2}}}$

Multiplying denominator & numerator by 4

$\sf{\implies\: h_{1} . h_{2} = \frac{4 × u^{4}sin^{2}\theta cos^{2} \theta}{16g^{2}}}$

$\sf{\implies\: h_{1} . h_{2} = \frac{ ( u^{2}2sin\theta cos\theta)^{2}}{16g^{2}}}$

$\sf{\implies\: h_{1} . h_{2} = \frac{ (u^{2}sin2\theta)^{2}}{16g^{2}}}$

$\sf{\implies\: h_{1} . h_{2} = \frac{1}{16} × \frac{ (u^{2}sin2\theta)^{2}}{g^{2}}}$

$\sf{\implies\: h_{1} . h_{2} = \frac{1}{16} × \left(\frac{ u^{2}sin2\theta}{g}\right)^{2}}$

$\sf{\implies\: h_{1} . h_{2} = \frac{ R^{2}}{16}}$

$\sf{\implies\: R^{2} = 16\:h_{1}.h_{2}}$

$\sf{\implies\: R = \sqrt{16\:h_{1}.h_{2}}}$

$\boxed{\large{\sf{\green{\therefore\: R = 4 \sqrt{\:h_{1}.h_{2}}}}}}$

Hence proved

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Note:

• $\large{\sf{\purple{\: R = \frac{u^{2}sin2\theta}{g}}}}$

• $\large{\sf{\purple{\: h_{max} = \frac{u^{2}sin^{2}\theta}{2g}}}}$

• Range is equal for angle θ and (90 - θ)

• 2sinAcosA = sin2A