ball A is dropped from a hot air balloon rising at a costant velocity of 14,7 m.s'1 at a height of 19,7 m above the ground.the b
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Answer:
v= 4055.08m/s
Explanation:
This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.
We know for definition that,
We must find the highest point and the lowest point to identify the change in energy, so
Point a)
The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.
That is to say that the energy of that object is equal to,
Point B )
We now use the average radius distance from the earth.
Then,
By the law of conservation of energy we know that,
clearing v,
Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s