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
Given that,
Height h=5m
g=10m/s2
Now, for first drop
s1=ut+12gt2
5=0+5t2
t=1sec
It means that the third drop leaves after one second of the first drop. Or, each drop leaves after every 0.5 sec
Now, distance covered by the second drop in 0.5 s
s2=ut+12gt2
s=0+12×10×0.5×0.5
s=1.25m
Therefore, the distance of the second drop above the ground
s=s1−s2
s=5−1.25
s=3.75m
Hence, this is the required solution