Solution 16
1. Extracting Gun Velocity ($u$)
We are given the maximum range of the shell from the gun (assumed stationary for this spec) is $R_{max} = 22.5$ m.
Since $R_{max} = \frac{u^2}{g}$, we can find $u$: $$ u = \sqrt{R_{max} \cdot g} = \sqrt{22.5 \times 10} = \sqrt{225} = 15 \text{ m/s} $$ The platform velocity is also $v = 15$ m/s.
We are given the maximum range of the shell from the gun (assumed stationary for this spec) is $R_{max} = 22.5$ m.
Since $R_{max} = \frac{u^2}{g}$, we can find $u$: $$ u = \sqrt{R_{max} \cdot g} = \sqrt{22.5 \times 10} = \sqrt{225} = 15 \text{ m/s} $$ The platform velocity is also $v = 15$ m/s.
2. Velocity Components relative to Ground
Let $\theta$ be the angle of projection relative to the horizontal (platform). $$ V_x = u \cos\theta + v = 15 \cos\theta + 15 $$ $$ V_y = u \sin\theta = 15 \sin\theta $$
Let $\theta$ be the angle of projection relative to the horizontal (platform). $$ V_x = u \cos\theta + v = 15 \cos\theta + 15 $$ $$ V_y = u \sin\theta = 15 \sin\theta $$
3. Maximizing Range
The horizontal range $R$ is given by: $$ R = \frac{2 V_x V_y}{g} = \frac{2 (15 \cos\theta + 15)(15 \sin\theta)}{10} $$ $$ R = \frac{2 \cdot 15^2 (1 + \cos\theta)\sin\theta}{10} = 45 (\sin\theta + \sin\theta \cos\theta) $$ $$ R(\theta) = 45 (\sin\theta + \frac{1}{2}\sin 2\theta) $$ To find maximum range, set $\frac{dR}{d\theta} = 0$: $$ \frac{dR}{d\theta} = 45 (\cos\theta + \cos 2\theta) = 0 $$ $$ \cos 2\theta + \cos\theta = 0 $$ Using $\cos 2\theta = 2\cos^2\theta – 1$: $$ 2\cos^2\theta + \cos\theta – 1 = 0 $$ Solving this quadratic equation for $\cos\theta$: $$ \cos\theta = \frac{-1 \pm \sqrt{1 – 4(2)(-1)}}{4} = \frac{-1 \pm 3}{4} $$ Possible values: $\frac{1}{2}$ or $-1$. Since the shell is fired forward/upward, $\cos\theta = 0.5$.
The horizontal range $R$ is given by: $$ R = \frac{2 V_x V_y}{g} = \frac{2 (15 \cos\theta + 15)(15 \sin\theta)}{10} $$ $$ R = \frac{2 \cdot 15^2 (1 + \cos\theta)\sin\theta}{10} = 45 (\sin\theta + \sin\theta \cos\theta) $$ $$ R(\theta) = 45 (\sin\theta + \frac{1}{2}\sin 2\theta) $$ To find maximum range, set $\frac{dR}{d\theta} = 0$: $$ \frac{dR}{d\theta} = 45 (\cos\theta + \cos 2\theta) = 0 $$ $$ \cos 2\theta + \cos\theta = 0 $$ Using $\cos 2\theta = 2\cos^2\theta – 1$: $$ 2\cos^2\theta + \cos\theta – 1 = 0 $$ Solving this quadratic equation for $\cos\theta$: $$ \cos\theta = \frac{-1 \pm \sqrt{1 – 4(2)(-1)}}{4} = \frac{-1 \pm 3}{4} $$ Possible values: $\frac{1}{2}$ or $-1$. Since the shell is fired forward/upward, $\cos\theta = 0.5$.
Answer:
$$ \cos\theta = \frac{1}{2} \implies \theta = 60^\circ $$ The gun should be aimed at an angle of 60° to the horizontal.
$$ \cos\theta = \frac{1}{2} \implies \theta = 60^\circ $$ The gun should be aimed at an angle of 60° to the horizontal.