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. $2h_1{h}_2$
• 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,
$h_1=\frac{u^2{\sin }^2\theta }{2g}$
$h_2=\frac{u^2{\sin }^2(90-\theta )}{2g}$
$h_1=\frac{u^2{\sin }^2\theta }{2g}$
$h_2=\frac{u^2{\cos }^2\theta }{2g}$
$h_1{h}_2=\frac{u^4{\sin }^2\theta {\cos }^2\theta }{4g^2}$
$\Rightarrow {\left(\frac{2u\sin \theta \cos \theta }{g}\right)}^2\times \frac{1}{4g}\times \frac{1}{4}$
$h_1{h}_2=R^2\frac{1}{16}$
$\Rightarrow R^2=16h_1{h}_2$
$R=4\left(\sqrt{h_1{h}_2}\right)$