Question

The range $R$ of a projectile is same when its maximum heights are $h_1$and $h_2$. What is the relation between $R$ and $h_1$ and $h_2$?

• A. $R=7\sqrt{h_1{h}_2}$
• B. $R=4\sqrt{h_1{h}_2}$
• C. $R=5\sqrt{h_1{h}_2}$
• D. $R=6\sqrt{h_1{h}_2}$

Answer 1

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

$\text{Step 1: Expression for maximum height and Range [Refer Fig.]}.$

Let $u$ be the initial velocity of the projectile and $\theta$ be the angle of projection.

Maximum height of the projectile:$h=\frac{u^2{\sin }^2\theta }{2g};$

Range of the projectile: $R=\frac{u^2\sin 2\theta }{g}$

$\text{Step 2: Relation between}{h}_1,h_2\&R$

For equal ranges, the body should be projected with angle $\theta$ and ${90}^o-\theta$ from the horizontal.

$[R=\frac{u^2\sin 2({90}^o-\theta )}{g}$$=\frac{u^2\sin ({180}^o-2\theta )}{g}$$=\frac{u^2\sin 2\theta }{g}]$

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

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

Multiplying $h_1$ and $h_2$

$h_1{h}_2=\frac{u^4{\sin }^2\theta {\cos }^2\theta }{4g^2}$$=\frac{1}{16}{\left(\frac{u^2\sin 2\theta }{g}\right)}^2$

$\Rightarrow$ $16h_1{h}_2=R^2$

$\Rightarrow$ $R=4\sqrt{h_1{h}_2}$

Hence, Option $B$ is correct.