Question

Water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at an instant when the first drop touches the ground. The height, in m, of the second drop above the ground at that instant is (Take g = 10 m s$^{-2}$)

• A. 1.25
• B. 2.5
• C. 3.75
• D. 5

Answer 1

Answer: (C)

3.75

Answer 2

Given, height of tap from ground,$h=.5m$
$g= 10\, m \,s^{-2}$
Time taken by first drop to reach the ground,
$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times5}{10}} = \sqrt{1} =1 s$
Now, if first drop falls at O s, and second drops at 1/2 s, then third drop will fall at 1 s as they fall on equal time interval.
At $t = 1\, s$, distance covered by 2$^{nd}$ drop is,
$h' =\frac{1}{2} gt^{'2} = \frac{1}{2}\left(10\right)\left(\frac{1}{2}\right)^{2} = \frac{10}{8} = 1.25 m$
Height of the second drop from the ground,
= $h- h'$
= 5 -1.25 = 3.75 m.