Kinematics of free fall — speed from a height A body is released from rest from a height h above the ground (neglect air resistance). What is its speed just before reaching the ground?

Introduction / Context:
Free-fall problems are fundamental in engineering mechanics. The relationship between displacement, acceleration due to gravity, and impact speed is used in impact, energy, and safety calculations.

Given Data / Assumptions:

• Initial speed u = 0 (released from rest).
• Downward constant acceleration a = g (use g ≈ 9.8 m/s^2).
• Vertical drop (displacement) s = h.

Concept / Approach:
Use the constant-acceleration kinematic equation without time: v^2 = u^2 + 2 a s. For free fall from rest, u = 0 and a = g. Alternatively, energy conservation gives the same result: loss in potential energy equals gain in kinetic energy.

Step-by-Step Solution:

Verification / Alternative check:
Energy method: m g h = (1/2) m v^2 → v = √(2 g h), consistent with kinematics and independent of mass m.

Why Other Options Are Wrong:

• √(g h): Missing factor 2.
• 2 g h or g h: Dimensions are incorrect for speed; those are acceleration*length.
• √(h/g): Has wrong dependence and dimensions.

Common Pitfalls:
Confusing velocity and speed or dropping the factor of 2 in the v^2 relation; using the wrong initial speed.

Final Answer:
v = √(2 g h)