Free fall from height h – find the impact speed at ground Neglecting air resistance, a body is dropped from rest from a height h above level ground. What is the correct expression for its speed v just before impact?
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Av = sqrt(2 g h)
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Bv = 2 g h
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Cv = sqrt(g / (2 h))
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Dv = (1/2) * g * h
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Ev = h^2 / (2 g)
Answer
Correct Answer: v = sqrt(2 g h)
Explanation
Introduction / Context:Impact speed from free fall is a fundamental kinematics result used across engineering—estimating drop tests, safety barriers, and potential energy conversion. It follows from constant-acceleration motion under gravity without air drag.
Given Data / Assumptions:
- Initial velocity u = 0 from height h.
- Constant acceleration a = g downward.
- No air resistance; motion is vertical.
Concept / Approach:
Use the kinematic energy–speed relation or the third equation of motion. Either approach removes time and relates velocity directly to displacement under constant acceleration.
Step-by-Step Solution:
From v^2 = u^2 + 2 a s with u = 0 and s = h: v^2 = 2 g h.Therefore v = sqrt(2 g h). Take the positive root for downward speed magnitude.Energy method cross-check: m g h converts to kinetic energy (1/2) m v^2 → v = sqrt(2 g h).Verification / Alternative check:
Dimensions: g has L/T^2 and h has L; 2 g h has L^2/T^2; its square root has L/T, the correct unit of speed.
Why Other Options Are Wrong:
(b) and (d) are missing the square root and have wrong units. (c) inverts the relationship. (e) has incorrect dependence on h and g.
Common Pitfalls:
Forgetting the square root; using time-dependent formulas unnecessarily; sign mistakes when taking displacement as negative.
Final Answer:
v = sqrt(2 g h)