Source: Principles of Physics
Problem Sets:
Problem
A projectile is launched from the ground with an initial speed $v_0$. To be successful, it must pass over a barrier of height $y$ located at a horizontal distance $x$ from the launch point.
- What is the minimum launch angle $\theta_{min}$ that will clear the barrier?
- What is the maximum launch angle $\theta_{max}$ that will clear the barrier?
[Q1] $\theta_{min} = \arctan\left( \frac{v_0^2 - \sqrt{v_0^4 - g(2yv_0^2 + gx^2)}}{gx} \right)$ [Q2] $\theta_{max} = \arctan\left( \frac{v_0^2 + \sqrt{v_0^4 - g(2yv_0^2 + gx^2)}}{gx} \right)$
The trajectory of the projectile is described by the equation:
$$y = x \tan(\theta) - \frac{gx^2}{2v_0^2 \cos^2(\theta)}$$Using the trigonometric identity $1/\cos^2(\theta) = 1 + \tan^2(\theta)$ and letting $T = \tan(\theta)$, we can form a quadratic equation in terms of $T$.
$$y = xT - \frac{gx^2}{2v_0^2}(1 + T^2)$$Rearranging into the standard form $AT^2 + BT + C = 0$:
$$ \left(\frac{gx^2}{2v_0^2}\right)T^2 - xT + \left(y + \frac{gx^2}{2v_0^2}\right) = 0 $$To simplify, multiply the entire equation by $\frac{2v_0^2}{gx^2}$:
$$ T^2 - \left(\frac{2v_0^2}{gx}\right)T + \left(1 + \frac{2v_0^2 y}{gx^2}\right) = 0 $$Solving for $T$ using the quadratic formula gives two possible values, which correspond to the minimum and maximum launch angles.
$$ T = \frac{1}{2} \left[ \frac{2v_0^2}{gx} \pm \sqrt{\left(\frac{2v_0^2}{gx}\right)^2 - 4\left(1 + \frac{2v_0^2 y}{gx^2}\right)} \right] $$ $$ T = \frac{v_0^2}{gx} \pm \frac{\sqrt{v_0^4 - g^2x^2 - 2gyv_0^2}}{gx} $$The two angles are then found by taking the arctangent of these two solutions for $T$. The smaller angle corresponds to the minus sign, and the larger angle corresponds to the plus sign.