Question

A projectile can have the same range for two angles of projection. If $h_1$ and $h_2$ are maximum heights when the range in the two cases is R, then the relation between R, $h_1$ and $h_2$ is

• A. $R=4\sqrt{h_1{h}_2}$
• B. $R=2\sqrt{h_1{h}_2}$
• C. $R=\sqrt{h_1{h}_2}$
• D. None of these

Answer 1

The correct option is A $R=4\sqrt{h_1{h}_2}$

$h_1=\frac{u^2{sin}^2\theta }{2g}$

$h_2=\frac{u^2{sin}^2\theta (90-\theta )}{2g},R=\frac{u^2{sin}^2\theta }{g}$

Range R is same for angle $\theta$ and $({90}^o-\theta )$

$\therefore h_1{h}_2=\frac{u^2{sin}^2\theta }{2g}\times \frac{u^2{sin}^2\theta (90-\theta )}{2g}$

$=\frac{u^4\left({sin}^2\theta \right)\times {sin}^2(90-\theta )}{4g^2}$

$[\because \sin (90-\theta )=\cos \theta ]$

$=\frac{u^4\left({sin}^2\theta \right)\times {cos}^2\theta }{4g^2}$

$[\because {sin}^2\theta =2\sin \theta \cos \theta ]$

$=\frac{u^4{\left(\sin \theta \cos \theta \right)}^2}{4g^2}=\frac{u^4{\left({sin}^2\theta \right)}^2}{16g^2}$

$=\frac{u^4{\left(\sin \theta \cos \theta \right)}^2}{16g^2}=\frac{R^2}{16}$

or, $R^2=16h_1{h}_{2}orR=4\sqrt{h_1{h}_2}$