Question

The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at the instant the first touches the ground. How far above the ground is the second drop at that instant?
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Answer 1

Answer:

The second drop is 3.75 m above the ground

Explanation:

Free Fall Motion

A free-falling object falls under the sole influence of gravity without air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is $g = 9.8 m/s^2$.

The distance traveled by a dropped object is:

$\displaystyle y=\frac{gt^2}{2}$

If we know the height h from which the object was dropped, we can find the time it takes fo hit the ground:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

When the first drop touches the ground there are two more drops in the air: the second drop still traveling, and the third drop just released from the tap.

The total time taken for the first drop to reach the ground is:

$\displaystyle t_1=\sqrt{\frac{2*5}{g}}$

$t_1 = 1.01\ s$

Half of this time has taken the second drop to fall:

$t_2 = 1.01\ s/2=0.505\ s$

It has fallen a distance of:

$\displaystyle y_2=\frac{9.8(0.505)^2}{2}$

$y_2 = 1.25\ m$

Thus its height is:

h = 5 - 1.25 = 3.75

The second drop is 3.75 m above the ground