A stalactite on the roof of a cave drips water at a steady rate to a pool 4.0 m below. As one drop of water hits the pool, a second drop is in the air, and a third is just detaching from the stalactite. (a) What are the position and velocity of the second drop when the first drop hits the pool? (b) How many drops per minute fall into the pool?
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We are required to calculate the position and velocity of the drop 2.
Consider the drop 1 first.
The separation between the stalactite and the pool is m.
From the equation,
When and , and ,
Substitute here and values,
The 3rd drop is about to leave the stalactite while the 1st drop hits the pool. The time taken by the first drop is 0.903 s to reach the pool. Therefore, the second drop would have traveled fors = 0.452 s.
The distance moved by the second drop after this time is m
Therefore, the second drop would be 1.00 m from the stalactite and 3.0 m above the pool.
The velocity of the second drop,
The velocity of the second drop at that instant would be 4.43 m/s.