Question

A canon can fire shells at speed $u$. Inclination of its barrel to the horizontal can be changed in steps of $\Delta \theta =1^{\circ }$ ranging from ${\theta }_1={15}^{\circ }to{\theta }_2={85}^{\circ }$. Let $R_n$ be the horizontal range for projection angle $\theta =n^{\circ }$. If $\Delta R_n=|R_n-R_{n+1}|$, the value of $n$ such that $\Delta R_n$ is maximum is
Neglect air resistance.

Answer 1

Range of a projectile, $R=\frac{u^{2}sin2\theta }{g}$
The graph of R vs $\theta$ is as shown.



In the context of the question,
the graph ranges from $A$ to $B$ and to $C$.
Slope of the graph is maximum at $C$.
[Graph is symmetrical about $\theta ={45}^{\circ },$ slope at $\theta ={15}^{\circ }$ is same as slope at ${75}^{\circ }$
It means $\frac{\Delta R}{\Delta \theta }$ is maximum at C. Hence answer is $\theta ={84}^{\circ }$.


(OR)

Range of a projectile, $R=\frac{u^{2}sin2\theta }{g}$
Range when projected at $n^{\circ }$ is $R_n=\frac{u^{2}sin2n}{g}$
Range when projected at $(n+1{)}^{\circ }$ is $R_{n+1}=\frac{u^{2}sin2(n+1)}{g}$

Now
$\Delta R_n=|R_n-R_{n+1}|$
$\Delta R_n=|\frac{u^{2}sin2n}{g}-\frac{u^{2}sin2(n+1)}{g}|$
$\Delta R_n=|\frac{u^2(sin2n-sin2(n+1\left)\right)}{g}|$
$\Delta R_n=|\frac{u^2(2sin\frac{(2n-2n-2)}{2}cos\frac{2n+2(n+1)}{2}}{g}|$
$\Delta R_n=|\frac{u^2\left(2sin\right(-1\left)cos\right(2n+1)}{g}|$
$\Delta R_n=\frac{u^2\left(2sin\right(1\left)|cos\right(2n+1)|}{g}$

As $n$ varies from ${15}^{\circ }to{84}^{\circ }$
$2n+1$ varies from ${31}^{\circ }to{169}^{\circ }$
$|cos(2n+1)|$ is maximum when $n={84}^{\circ }$