An airplane flies at a constant height h and constant speed j. A gun is at the origin (0,0). A shell is fired from the gun with an initial muzzle velocity v at an angle theta with the horizontal. What

The correct option is C. $v = (j^2 + 2gh)^{1/2}$.

Step 1: Define the motion of the airplane and the shell.
Let the gun be at the origin $(0,0)$.
The airplane flies at a constant height $h$ and constant speed $j$. At time $t$, its position is $(x_A, y_A) = (jt, h)$.
The shell is fired with an initial muzzle velocity $v$ at an angle $\theta$ with the horizontal. At time $t$, its position is $(x_S, y_S)$.
The horizontal component of the shell's initial velocity is $v_x = v \cos\theta$.
The vertical component of the shell's initial velocity is $v_y = v \sin\theta$.
The equations of motion for the shell are:

$$x_S(t) = (v \cos\theta) t$$ $$y_S(t) = (v \sin\theta) t - \frac{1}{2} g t^2$$

Step 2: Set the conditions for the shell to hit the airplane.
For the shell to hit the airplane at time $t$, their positions must be equal:

$$x_S(t) = x_A(t) \implies (v \cos\theta) t = j t$$

Since $t \neq 0$ (the shell takes time to reach the airplane), we can divide by $t$:

$$v \cos\theta = j \quad (*)$$ $$y_S(t) = y_A(t) \implies (v \sin\theta) t - \frac{1}{2} g t^2 = h \quad (**)$$

Step 3: Express $\sin\theta$ in terms of $v$ and $j$.
From $(*)$, $\cos\theta = \frac{j}{v}$.
Using the identity $\sin^2\theta + \cos^2\theta = 1$:

$$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \left(\frac{j}{v}\right)^2} = \sqrt{\frac{v^2 - j^2}{v^2}} = \frac{\sqrt{v^2 - j^2}}{v}$$

Step 4: Substitute $\sin\theta$ into equation $(**)$ and solve for $t$.
Substitute $\sin\theta$ into $(**)$:

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