Question

A boy throws a ball with an initial velocity and at an angle of elevation such that the range is $r$ and maximum height to which ball rises is $h$. Find the maximum range that can be obtained with same initial velocity.

• A. $h+\frac{r^2}{8h}$
• B. $2h+\frac{r^2}{8h}$
• C. $2h-\frac{r^2}{8h}$
• D. $h-\frac{r^2}{8h}$

Answer 1

The correct option is B $2h+\frac{r^2}{8h}$

The maximum height of a projectile having initial velocity $u$ and projection angle $\alpha$ is given by, $$h=\frac{u^2{\sin }^2\alpha }{2g}....\left(1\right)$$And range, $$r=\frac{u^2\sin 2\alpha }{g}...\left(2\right)$$ Maximum range of the projectile at $\alpha ={45}^{\circ }$ is $$R_m=\frac{u^2}{g}...\left(3\right)$$
Now substituting maximum range $R_m$ in $\left(1\right)$ and $\left(2\right)$, we get

$2h=R_m{\sin }^2\alpha ...\left(4\right)$

$r=R_m\sin 2\alpha ...\left(5\right)$

Equation $\left(4\right)$ can be rewritten in the form,

$h=\frac{R_m}{4}(1-\cos 2\alpha )$

$\Rightarrow \cos 2\alpha =1-\frac{4h}{R_m}...\left(6\right)$

Equation $\left(5\right)$ can be rewritten in the form

$\sin 2\alpha =\frac{r}{R_m}...\left(7\right)$
Now,

${\sin }^{2}2\alpha +{\cos }^{2}2\alpha =1$

Using equation $\left(6\right)$ and $\left(7\right)$,

$\Rightarrow {\left(\frac{r}{R_m}\right)}^2+{(1-\frac{4h}{R_m})}^2=1$

$\Rightarrow \frac{r^2}{R_m^2}+1+\frac{16h^2}{R_m^2}-\frac{8h}{R_m}=1$

$\Rightarrow \frac{r^2}{R_m^2}+\frac{16h^2}{R_m^2}-\frac{8h}{R_m}=0$

Multiplying both side by $R_m^2$

$\Rightarrow r^2+16h^2-8h{R}_m=0$

$\Rightarrow R_m=2h+\frac{r^2}{8h}$

Hence, option (b) is the correct answer.

Alternate solution:

We know that, $r=\frac{4h}{\tan \alpha }$

Squaring on both sides, we get

$r^2=\frac{16h^2}{{\tan }^2\alpha }$

$\Rightarrow r^2=\frac{16h^2\tan 2\alpha }{\tan 2\alpha -2\tan \alpha }$

From $\left(6\right)$ and $\left(7\right)$

$r^2=\frac{16h^2}{1-\frac{8h(R_m-4h)}{r^2}}$

$\Rightarrow r^2-8h{R}_m+32h^2=16h^2$

$\Rightarrow R_m=2h+\frac{r^2}{8h}$