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A climber using bottled oxygen accidentally drops the oxygen bottle from an altitude of 4500 m. If the bottle fell straight down this entire distance, what is the velocity of the 3-kg bottle just prior to impact at sea level? (Note: ignore air resistance)

User Sgliser
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1 Answer

3 votes

Answer:

300 m/s

Step-by-step explanation:

As the cylinder drops off so its initial velocity is zero.

h = 4500 m, g = 10 m/s^2, u = 0

Use third equation of motion

v^2 = u^2 + 2 g h

v^2 = 0 + 2 x 10 x 4500

v^2 = 90000

v = 300 m /s

User Achref
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