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.
1 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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