Problem: Water drops from a water tap on to the floor x m. below. The first drop strikes the floor at the instant the fourth one begins to fall. Determine the positions of the drops when a drop just strikes the floor. Drops fall at regular intervals of time.

Solution:
To solve the problem, we need to use the equations of motion for falling objects. Let us assume that the time interval between the drops is t seconds. Then, the position of the nth drop at any time t is given by the equation:

s_n = ut + (1/2)at^2

where u is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2), and s_n is the position of the nth drop.

Step 1: Find the time interval between the drops
Since the drops fall at regular intervals of time, we can assume that the time interval between the drops is the same. Let this time interval be t seconds.

Step 2: Find the position of the fourth drop
According to the problem statement, the first drop strikes the floor at the instant the fourth one begins to fall. Therefore, the time taken by the first three drops to fall from the tap to the floor is equal to the time taken by the fourth drop to fall from the tap to the floor.

Let us assume that the position of the fourth drop when it starts to fall is s_4. Then, using the equation of motion, we have:

s_4 = (1/2)at^2

Step 3: Find the position of the other drops when they strike the floor
Now that we know the position of the fourth drop when it starts to fall, we can find the position of the other drops when they strike the floor.

Let us assume that the position of the nth drop when it strikes the floor is s_n'. Then, using the equation of motion, we have:

s_n' = s_4 + (n-4)gt

where g is the acceleration due to gravity (9.8 m/s^2) and t is the time interval between the drops.

Step 4: Find the value of t
The value of t can be found by using the fact that the first drop strikes the floor at the same time as the fourth drop begins to fall. Therefore, the time taken by the first three drops to fall from the tap to the floor is equal to the time taken by the fourth drop to fall from the tap to the floor. Using the equation of motion, we have:

x = (1/2)at^2

Solving for t, we get:

t = sqrt(2x/a)

Step 5: Substitute the values and find the positions of the drops
Substituting the values of s_4, g, and t in the equation for s_n', we get:

s_n' = (1/2)at^2 + (n-4)gt

s_n' = (1/2)(9.8)(sqrt(2x/9.8))^2 + (n-4)(9.8)(sqrt(2x/9.8))

Simplifying, we get:

s_n' = x + (n-4)(sqrt(2ax))

Therefore,

Let the 4 drops be D1, D2, D3, D4Consider a drop falling from tap. WKT s=ut+(1/2)at(t)Here s=x, u=0,a=g.x=0t+(1/2)gt(t)x=(1/2)gt(t)t is time taken by drop to reach ground from tap. t=square root of (2x/g)Let time interval between two successive by drop falls be T. There are 4drops so there are 3T's in one cycle. 3T=tT=t/3=( root of 2x/g)/3=root of (2x/9g)Now when D1 hits ground and D4 starts out D3:- D3 spends T time since its release from tap. Let s be distance between D3 and taps=(1/2)gT(T)s=(1/2)g(2x/9g)s=x/9Distance from ground =x-x/9=8x/9D2:- D2 spends 2T time since its releases=(1/2)g(2T)(2T)s=(1/2)g(4)(2x/9g)s=4x/9Distance from ground=x-4x/9=5x/9