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 case $R,$ 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 $R=4\sqrt{h_1{h}_2}$

$h_1=\frac{u^2{\sin }^2\theta }{2g}$
$h_2=\frac{u^2{\sin }^2(90-\theta )}{2g},R=\frac{u^2\sin 2\theta }{g}$

Range $R$ is same for angle $\theta and(90-\theta )$
$\therefore h_1{h}_2=\frac{u^2{\sin }^2\theta }{2g}\times \frac{u^2{\sin }^2(90-\theta )}{2g}$
$=\frac{u^4\left({\sin }^2\theta \right)\times {\sin }^2(90=\theta )}{4g^2}$

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

$\frac{u^4(\sin \theta \cos \theta {)}^2}{4g^2}=\frac{u^4(sin2\theta {)}^2}{16g^2}$

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

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