Question

A stone projected from ground with certain speed at an angle $\theta$ with horizontal attains maximum height $h_{1}$. When it is projected with same speed at an angle $\theta$ with vertical attains height $h_{2}$. The horizontal range of projectile is

• A. $\frac{h_{1}+h_{2}}{2}$
• B. $2 h_{1} h_{2}$
• C. $4 \sqrt{h_{1} h_{2}}$
• D. $h_{1}+h_{2}$

Answer 1

Answer: (C)

When the angles are complimentary the range is same,
$h_{1}=\frac{u^{2} \sin ^{2} \theta}{2 g}$
$h_{2}=\frac{u^{2} \sin ^{2}(90-\theta)}{2 g}$
$h_{1}=\frac{u^{2} \sin ^{2} \theta}{2 g}$
$h_{2}=\frac{u^{2} \cos ^{2} \theta}{2 g}$
$h_{1} h_{2}=\frac{u^{4} \sin ^{2} \theta \cos ^{2} \theta}{4 g^{2}}$
$\Rightarrow \left(\frac{2 u \sin \theta \cos \theta}{g}\right)^{2} \times \frac{1}{4 g} \times \frac{1}{4}$
$h_{1} h_{2}=R^{2} \frac{1}{16}$
$\Rightarrow R^{2}=16\, h_{1} h_{2}$
$R=4\left(\sqrt{h_{1} h_{2}}\right)$