An airplane is flying overhead at a constant elevation of 3000ft. A man is viewing the plane from a position 4000ft from the bas
e of a radio tower. The airplane is flying horizontally away from the man. If the plane is flying at a rate of 500ft/s, at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower?
1 answer:
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Answer:
Emechanical=mgh+mν²
Explanation:
The equation for the total mechanical energy is:
Emechanical=Epotential+Ekinetic
In which,
Epotential=mgh; m: mass of the body, g: gravity; h: height
Ekinetic=mν²; m: mass of the body, ν: velocity of the body
So,
Emechanical=mgh+mν²