a tap of a height of a 5m from the ground is leaking water drop at regular intervals.the fourth water drop leaves the tap at the instant the first hit the ground. find the height of the second drop from the ground at the instant.
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Answer (1)
14th Jan, 2021
Dear student,
Height of tap from ground = 5 m, let's assume g = 10m/s^2 for ease calculation.
Now, time taken by first drop to fall to ground :-
5 = ut + 1/2 gt^2
5 = 0 + 1/2 10 t^2 (because u=0)
Therefore t = 1 sec, i.e, first drop takes 1 second to reach the ground.
Now, water drops are falling at regular intervals.
So, as fourth drop immediately when the first drop reaches ground, so for each drop the time interval will be of 1/3 seconds.
Now, when the first drop falls on ground the second drop travels for 2/3 seconds.
So, the distance traveled by the second from the tap :-
s = ut + 1/2 gt^2
s = 1/2 10 (2/3)^2 = 20/9 m
So, distance of second drop from the ground = 5 - 20/9 = 25/9 =2.778m
Height of tap from ground = 5 m, let's assume g = 10m/s^2 for ease calculation.
Now, time taken by first drop to fall to ground :-
5 = ut + 1/2 gt^2
5 = 0 + 1/2 10 t^2 (because u=0)
Therefore t = 1 sec, i.e, first drop takes 1 second to reach the ground.
Now, water drops are falling at regular intervals.
So, as fourth drop immediately when the first drop reaches ground, so for each drop the time interval will be of 1/3 seconds.
Now, when the first drop falls on ground the second drop travels for 2/3 seconds.
So, the distance traveled by the second from the tap :-
s = ut + 1/2 gt^2
s = 1/2 10 (2/3)^2 = 20/9 m
So, distance of second drop from the ground = 5 - 20/9 = 25/9 =2.778m
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Comments (3)
thx bro
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goldenrazgoldenrz17045
bro agr wha pr 3rd hota to
14th Jan, 2021
goldenrazgoldenrz17045
If the third drop falls just after fall of 1st drop on ground, then the time interval will be of 1/2 seconds.
