Question

Water drips from a faucet at the rate of two drops per second. What distance separates one drop from the following drop 0.75 seconds after the leading drop leaves the​ faucet? How much time elapses between impacts of the two drops if they fall onto a surface that is 3 feet below the lip of the​ faucet?

Answer 1

So here it is given that the drops fall at a rate of 2 drops per second, thus we can say that the drops fall from the faucet at an interval of 0.5 seconds.

Falling of the drop from the faucet is an example of uniform accelerated motion with initial velocity being zero(as it is at rest at the faucet at time t = 0 seconds).

Now let us take the case of first drop, Let us calculate how much distance the drop ca travel in the given 0.75 seconds,

so using the uniform acceleration equation consisting of distance and time

           $d= ut+(\frac{1}{2}at^{^{2}})$

here u = initial velocity which is zero in this case.

    d₁ = distance covered by first droplet in 0.75 seconds

    t = time taken to travel = (0.75 seconds for first drop and 0.25 seconds for second drop)

    a = g = 9.81 m/s²

therefore the equation becomes

     $d_{1}= \frac{1}{2}\times 9.81\times 0.75^{2}= 2.75 m$

Now for the second droplet, which will be dropped after 0.5 seconds after the first droplet, it will have 0.25 seconds to travel (to complete the 0.75 seconds target time)

    d₂ = distance covered by the second droplet in 0.25 seconds

using the same equation of uniform acceleration for the second droplet condition, we will get,

     $d_{2}= \frac{1}{2}\times 9.81\times 0.25^{2}= 0.30 m$

Therefore the distance that separates the two droplets at 0.75 seconds will be the difference of the distance covered by both the droplets,

therefore, d = Distance between the two droplets at time 0.75 seconds

                    = d₁ - d₂ = (2.75 - 0.3) m = 2.45m

Now for the second part of the problem, if the floor is at the distance of 0.91 m (3 feet) away from the faucet lip, then for finding the time between two successive droplets impacts on the floor can be found in the following way:

Let us consider that t₁ = time taken in seconds for the first droplet to hit the floor.

here in this case also the initial velocity of the drop is zero, as it is at rest before dropping from the faucet.

Then using the same uniform acceleration equation, this time with a known distance of 0.91m (3 feet) we get,

      $0.91 = \frac{1}{2}\times 9.81\times t_{1}^{2}$

    

which means that for the first drop it takes 0.43 seconds to hit the floor, which means that the second droplet has still 0.07 seconds to fall off from the faucet (as the two droplets fall at an interval of 0.5 seconds).

Once the second droplet falls off the faucet it will take the same amount of time to reach the floor as taken by the first droplet, that is 0.43 seconds.

Thus the answer to the question; how much time elapsed between the impact of two drops can be given as

Total time, t = t₁ + 0.07 seconds = 0.43 + 0.07 = 0.5 seconds