Question

Water drops fall at regular intervals from a tap 5 m above ground. The 3 rd drop is leaving tap when first drop reaches ground. The distance of 2 nd drop at the instant is

• A. 2.5 m
• B. 3.75 m
• C. 4 m
• D. 1.25 m

Answer 1

Answer: (B)

Given that,

Height $h=5m$

$g=10m/s^2$

Now, for first drop

$s_1=ut+\frac{1}{2}g{t}^2$

$5=0+5t^2$

$t=1\sec$

It means that the third drop leaves after one second of the first drop. Or, each drop leaves after every $0.5sec$

Now, distance covered by the second drop in $0.5s$

$s_2=ut+\frac{1}{2}g{t}^2$

$s=0+\frac{1}{2}\times 10\times 0.5\times 0.5$

$s=1.25m$

Therefore, the distance of the second drop above the ground

$s=s_1-s_2$

$s=5-1.25$

$s=3.75m$

Hence, this is the required solution