Source: High school physics (Chinese)
Problem
An airplane flies horizontally at a speed $v_1$ at an altitude $h$ above the ground. When the airplane is directly above an anti-aircraft gun, the gun fires a shell with an initial velocity $v_0$ at an angle $\alpha$ with the horizontal. Air resistance is negligible.
[Q1] For the shell to hit the plane, two conditions must be met simultaneously: the horizontal velocities must match ($v_1 = v_0 \cos\alpha$), and the shell must be able to reach the plane's altitude $h$. The second condition requires the time-of-flight equation, $h = (v_0 \sin\alpha)t - \frac{1}{2}gt^2$, to have a real solution for $t$, which implies its discriminant must be non-negative: $(v_0\sin\alpha)^2 \ge 2gh$. Using the identity $\sin^2\alpha = 1 - \cos^2\alpha$ and substituting $\cos\alpha = v_1/v_0$, we get:
$$v_0^2(1 - (v_1/v_0)^2) \ge 2gh$$ $$v_0^2 - v_1^2 \ge 2gh$$Therefore, the condition is:
$$v_0^2 \ge v_1^2 + 2gh$$Let the origin (0, 0) be the location of the anti-aircraft gun. The airplane flies along the line $y=h$. At time $t=0$, the plane is at $(0, h)$ and the shell is fired from $(0, 0)$.
The position of the airplane at time $t$ is given by:
$$x_p(t) = v_1 t$$ $$y_p(t) = h$$The position of the shell at time $t$ is given by:
$$x_s(t) = (v_0 \cos\alpha) t$$ $$y_s(t) = (v_0 \sin\alpha) t - \frac{1}{2}gt^2$$For the shell to hit the airplane, there must be a time $t > 0$ where their positions coincide, i.e., $x_p(t) = x_s(t)$ and $y_p(t) = y_s(t)$.
1. Condition from Horizontal Motion Equating the x-coordinates:
$$v_1 t = (v_0 \cos\alpha) t$$Since $t > 0$, we can divide by $t$ to find the condition on the launch angle $\alpha$:
$$v_1 = v_0 \cos\alpha \quad \implies \quad \cos\alpha = \frac{v_1}{v_0}$$2. Condition from Vertical Motion Equating the y-coordinates:
$$h = (v_0 \sin\alpha) t - \frac{1}{2}gt^2$$This can be rearranged into a quadratic equation for the time of impact, $t$:
$$\frac{1}{2}gt^2 - (v_0 \sin\alpha)t + h = 0$$For a real solution for $t$ to exist (i.e., for the shell to be able to reach the altitude $h$), the discriminant of this quadratic equation must be non-negative ($\Delta = b^2 - 4ac \ge 0$).
$$\Delta = (-v_0 \sin\alpha)^2 - 4\left(\frac{1}{2}g\right)(h) \ge 0$$ $$(v_0 \sin\alpha)^2 - 2gh \ge 0$$ $$v_0^2 \sin^2\alpha \ge 2gh$$3. Combining Conditions To find the overall condition on the initial velocity $v_0$, we eliminate the angle $\alpha$ using the trigonometric identity $\sin^2\alpha + \cos^2\alpha = 1$, which gives $\sin^2\alpha = 1 - \cos^2\alpha$.
Substitute $\cos\alpha = v_1/v_0$ into the identity:
$$\sin^2\alpha = 1 - \left(\frac{v_1}{v_0}\right)^2 = \frac{v_0^2 - v_1^2}{v_0^2}$$Now substitute this expression for $\sin^2\alpha$ into the inequality from the vertical motion:
$$v_0^2 \left(\frac{v_0^2 - v_1^2}{v_0^2}\right) \ge 2gh$$ $$v_0^2 - v_1^2 \ge 2gh$$Finally, rearranging the terms gives the required condition for a hit to be possible:
$$v_0^2 \ge v_1^2 + 2gh$$This condition ensures that there exists a real launch angle $\alpha$ for which the shell can intercept the plane.