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Question

a tap drops water bubbles in same time intervals.when the bubble drop"B" drops from the tap ,bubble drop "A" 0.25m away from the tap(drops fall virticaly).count the distance of bubble "A"(from the tap) when the distance between "A" & "B" are 0.75m
g=10ms^-2

Answer 1

Using h=ut +1/2 gt^2
For DROP-A
h1=0(t1) +0.5(10)(t1)^2........i
0.25=5(t1)^2
t1^2=0.25/5=0.05.............ii
For DROP-B
h2=0(t2) +0.5(10)(t2)^2
h2=5(t2)^2......................iii
h1-h2=5(t1+t2)^2-5(t2)^2
but h1-h2=0.75 and (t1)^2=0.05
0.75=5{(t1^2+2t1*t2 +t2^2)-t1^2}
0.75=5{2t1*t2 +t2^2)}
0.75=5{2(0.223606797)t2 +t2^2)}
0.75=2.24t2 +5t2^2
5t2^2+2.24t2-0.75=0 solve for t2
t2={-2.24+/-(4.474)}/10
t2=0.25 seconds
h=0(0.25+0.22)+0.5(10)(0.47)^2
=1.1045 meters

Answer 2

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