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? A. 300 m/s. B. 150 m/s. C. 30 m/s. D. 3 m/s.

v = 300 m/s

Explanation:

Given that

altitude, h = 4500 m

The mass, m = 3 kg

Lets take acceleration due to gravity, g = 10 m/s²

The speed before impact at sea level = v

Initial speed, u = 0 m/s

We know that

v²=u²+2 g h

v=final speed

u=initial speed

h=height

Now by putting the values in the above equation

v² = 0² + 2 x 10 x 4500

v²=90000

v = 300 m/s

Therefore the speed at sea level will be 300 m/s.