A drop of water falls from a height of 200 m. What is the difference in the distance traveled by the drop in the 4th and 8th second after it falls. (Use g=9.8 m/s^2)

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The drop of water falls from a height of 200 m. As a result the initial velocity of the drop is equal to 0 m/s. The acceleration due to the gravitational force of attraction of the Earth is equal to 9.8 m/s^2.

The distance traveled by a particle in time t if the initial velocity is u and the acceleration is a, is given by D = u*t + (1/2)*a*t^2

In the problem, it should be noted that the total time taken by the drop to fall 200 m is only `sqrt(200/4.9) ~~ 6.38` seconds.

To determine the difference in the distance traveled by the drop in the 8th second and the 4th second it is necessary to assume that it falls from a higher point.

The distance traveled by the drop of water in the 4th second is `D_4 - D_3` = 0*4 + (1/2)*9.8*4^2 - 0*3 - (1/2)*9.8*3^2 = 4.9*(16 - 9) = 34.3 m

The distance traveled by the particle in the 8th second is `D_8 - D_7` = 0*8 + (1/2)*9.8*8^2 - 0*7 - (1/2)*9.8*7^2 = 4.9*(64 - 49) = 73.5 m

The difference between the two is 39.2 m.

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