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

Answer :

Answer:

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.

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