## Kinematics Physics

The water drops are falling at regular intervals from a tap, which is 5 m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?

18-Nov-2015 6:42 PM
~6

1

GIven height: 5m

Time taken by the 1st drop to reach the ground:

S = ut + $\frac{1}{2}$at2

u=0, S=-5, a=-10

=> t = 1 sec.

Now, the third drop is just leaving the tap, the 2nd drop is somewhere midway, and the third drop is hitting the ground. And the time interval for this situation to occur is 1 sec.

Since it says the drops are falling at regular intervals, therefore, the time must be equally divided in that time interval of 1 sec, giving $\frac{1}{2}$ second of intervals between the drops. (meaning, the time interval between the first drop and second drop is supposed to be equal to the time interval between 2nd drop and 3rd drop (as drops leave at regular intervals) . So if total time is 1sec, dividing it into 2 intervals (1st drop and 2nd drop; 2nd drop and 3rd drop) will give 1/2 seconds, as the interval between 2 drops.

Now, the time taken by the second drop is 1/2 second. Calculate its distance as:

S = ut + $\frac{1}{2}$at2   (u=0, t=1/2)

S = 1.25

Distance from ground:

5 - 1.25

=> 3.75 m

21-11-2015 15:57
~32

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