Water droplets are falling from a barrel to the ground at regular intervals. The height between the barrel and the ground is 9m. When the first drop touches the ground the fourth drop just starts falling from the barrel. Calculate the height of the second and third drop of water from the barrel.
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4m and 1m
My attempt: Let the interval between the falling of two drops be "t" seconds.
Then 4th drop falls after 3t seconds of falling of 1st drop. Therefore, the time of flight of 1st drop is 3t seconds.
H=1/2 gt²
9=5*t²
t=3/√5
this is the time taken for 1st drop
therefore the time for the 3rd drop is 1/3*3/√5
From this equation, t'=√(1/5) seconds.
Distance of 2nd drop from the roof = H=1/2gt'²
H=1/2*10*1/5=1m
and for the 2nd drop the taken will be 2*1/√5
therefore, t''=2/√5
Distance of 3rd drop from the roof = H=1/2gt''²=1/2*10*4/5= 4m
Hence the answer is 1m and 4m
My attempt: Let the interval between the falling of two drops be "t" seconds.
Then 4th drop falls after 3t seconds of falling of 1st drop. Therefore, the time of flight of 1st drop is 3t seconds.
H=1/2 gt²
9=5*t²
t=3/√5
this is the time taken for 1st drop
therefore the time for the 3rd drop is 1/3*3/√5
From this equation, t'=√(1/5) seconds.
Distance of 2nd drop from the roof = H=1/2gt'²
H=1/2*10*1/5=1m
and for the 2nd drop the taken will be 2*1/√5
therefore, t''=2/√5
Distance of 3rd drop from the roof = H=1/2gt''²=1/2*10*4/5= 4m
Hence the answer is 1m and 4m
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0
Answer is 4m and 1m respectively
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