Question

On seeing enemy ships at a distance of 125$\sqrt{3}$ m from the shore the cannons start firing fireballs. If the cannons fire at a speed of 50 m/s then at what angle should they aim so they do not miss. Also at what angle can the cannons achieve the maximum distance?

• A. Angle to destroy ships = ${45}^{\circ }$; Angle for maximum range = ${90}^{\circ }$
• B. Angle to destroy ships = ${30}^{\circ }$; Angle for maximum range = ${45}^{\circ }$
• C. Angle to destroy ships = ${60}^{\circ }$; Angle for maximum range = ${45}^{\circ }$
• D. Angle to destroy ships = ${90}^{\circ }$; Angle for maximum range = ${90}^{\circ }$

Answer 1

Answer: (B), (C)

Angle to destroy ships = ${60}^{\circ }$; Angle for maximum range = ${45}^{\circ }$

Range or $s_x=\text{/}{125}_m\sqrt{3}=\frac{2\times \text{/}{2500}^{20}sin\theta cos\theta }{10}$
$\Rightarrow$ sin $\theta$ cos $\theta$ = $\frac{\sqrt{3}}{4}$
cos $\theta$ = $\frac{\sqrt{3}}{2}$ cos $\theta$ = $\frac{1}{2}$
$\Rightarrow$ $\theta$ can be either ${30}^{\circ }$ or ${60}^{\circ }$ Now to find the maximum distance that the cannon's can hit, we will have to maximize the range of cannons.
Range = $\frac{2\times 2500sin\theta cos\theta }{10}$
= 250 sin 2$\theta$ ( $\because$2 sin $\theta$ cos $\theta$ = sin 2$\theta$)
To maximize this we need to maximize sin 2$\theta$
We know that is 1
sin 2$\theta$ can be maximum 1
When 2$\theta$ = ${90}^{\circ }$
Or $\theta$ = ${45}^{\circ }$
So maximum Range = 250m.