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Asked by surabhilakhara7299 | 17 Jun, 2021, 03:54: PM
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Time t taken for the first drop to reach ground is determined from the following equation
 
h = (1/2) g t2 
 
where h = 9 m , is the vertical distance travelled by the first drop and g is acceleration due to gravity 
 
begin mathsize 14px style t space equals space square root of fraction numerator 2 space h over denominator g end fraction end root space equals space square root of fraction numerator 2 space cross times space 9 over denominator 9.8 end fraction end root space equals space 1.355 space s end style
Between 4 drops , there are three equal intervals Δt. 
 
Δt = ( 1.355 ) /3 = 0.452 s
 
Distance d3 of third drop from top of building is given as
 
begin mathsize 14px style d subscript 3 space equals space left parenthesis 1 divided by 2 right parenthesis space g space open parentheses 2 increment t close parentheses squared space equals space 0.5 space cross times 9.8 space cross times left parenthesis space 2 space cross times space 0.452 space right parenthesis squared space almost equal to 4 space m end style
Distance d2 of second drop from top of building is given as
 
begin mathsize 14px style d subscript 2 space equals space left parenthesis 1 divided by 2 right parenthesis space g space open parentheses increment t close parentheses squared space equals space 0.5 space cross times 9.8 space cross times left parenthesis space space 0.452 space right parenthesis squared space almost equal to 1 space m end style
Answered by Thiyagarajan K | 17 Jun, 2021, 04:58: PM
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