The water drop falls at regular intervals from a tap 9 m above the ground. The fourth drop is leaving the tap at the instant, the first drop touches the ground. How high is the third drop at that instant?

Asked by anujha54 | 13th May, 2021, 12:03: PM

Expert Answer:

Time taken t1 by first drop to travel through 9 m vertical distance is determined by the following equation
 
h1 = (1/2) g t12
 
where h1 = 9 m is vertical distance , g is acceleration due to gravity and t1 is time taken
 
begin mathsize 14px style t subscript 1 space equals space square root of fraction numerator 2 h subscript 1 over denominator g end fraction end root space equals space square root of fraction numerator 2 cross times 9 over denominator 9.8 end fraction end root space equals space 1.355 space s end style
There are three equal intervals when four drops are falling from tap in this time t = 1.355 s
 
interval time = ( 1.355 / 3 ) = 0.452 s
 
Hence third drop has travelled one interval times , i.e., t3 = 0.452 s
 
Vertical distance travelled by third drop, h3 = (1/2) g t32 = 0.5 × 9.8 × 0.452 × 0.452 = 1 m 
 
Hence third drop is at a height of 8 m above ground , when first drop has reached the ground

Answered by Thiyagarajan K | 13th May, 2021, 12:47: PM