Method: Tilted Frame Vector Analysis
We analyze the motion in a coordinate system attached to the box. Let the floor of the box be the $x’$-axis. This simplifies the problem by treating the “tilted floor” as our horizontal reference.
In this tilted frame (rotated by angle $\theta$), the gravity vector $\vec{g}$ is resolved into two components:
- Perpendicular to floor: $g_{\perp} = g \cos \theta$. (This component acts “downward” relative to the box floor and limits the jump height).
- Parallel to floor: $g_{\parallel} = g \sin \theta$.
To jump out of the box, the grasshopper must clear the top edge (height $h$) measured perpendicular to the floor.
The maximum vertical height $H’$ in this specific reference frame is determined purely by the perpendicular component of velocity ($u_{\perp}$) and the perpendicular gravity ($g_{\perp}$).
$$ H’_{max} = \frac{u_{\perp}^2}{2 g_{\perp}} $$
To minimize the required tilt (or maximize the ability to escape), the grasshopper should direct its entire initial velocity $u$ perpendicular to the floor. Thus, we set $u_{\perp} = u$.
Substitute $u_{\perp} = u$ and $g_{\perp} = g \cos \theta$ into the height equation: $$ H’_{max} = \frac{u^2}{2 g \cos \theta} $$ For the grasshopper to escape, this maximum height must be at least the height of the box $h$: $$ \frac{u^2}{2 g \cos \theta} \ge h $$ $$ \cos \theta \le \frac{u^2}{2gh} $$ Substituting the given values ($u=3, h=0.52, g=10$): $$ \cos \theta = \frac{3^2}{2(10)(0.52)} = \frac{9}{10.4} \approx 0.865 $$