Projectile in a low-height horizontal tunnel: A particle is fired inside a tunnel of clear height 554 cm with initial speed 60 m/s. To achieve maximum horizontal range without striking the roof, what should be the angle of projection (degrees from horizontal)?

Introduction / Context:

This problem blends projectile kinematics with a geometric constraint: the projectile must not exceed the tunnel roof height. The usual 45° angle that maximizes range in open space is impossible here because the peak height would be far above the tunnel ceiling. Therefore, we optimize the range subject to a maximum-height constraint.

Given Data / Assumptions:

• Tunnel height H = 554 cm = 5.54 m.
• Initial speed u = 60 m/s.
• Standard gravity g = 9.81 m/s^2.
• No air resistance; ground and roof are horizontal.

Concept / Approach:

The trajectory peak height for a projectile at angle theta is H_max = u^2 * sin^2(theta) / (2 * g). To avoid striking the roof, we require H_max <= H. For maximum allowed range under this constraint, the optimal angle is the largest theta that still satisfies the height limit (otherwise we could slightly increase theta and gain more range without hitting the roof).

Step-by-Step Solution:

Verification / Alternative check:

Open-terrain maximum range at 45° would have H_max = u^2 / (4 * g) = 3600 / 39.24 ≈ 91.8 m >> 5.54 m, confirming the roof is the active constraint. With theta ≈ 10°, the projectile just clears the roof at the apex and thus gives the largest possible range within the tunnel.

Why Other Options Are Wrong:

• 8°–9° are feasible but give a lower range than the constraint-limiting angle.
• 11°–12° would exceed the allowed peak height and strike the roof.

Common Pitfalls:

• Maximizing range using 45° without checking height restrictions.
• Using degrees vs. radians inconsistently in calculator steps.

Final Answer:

10°