A drop of water falls from a ledge of a tall building from a height 320 m. What is the difference between the distance traveled by the drop in the 1st second and the 3rd second after the fall?
The drop of water falls from a ledge of a tall building from a height 320 m.
An object initially traveling at speed u m/s and accelerating at a m/s^2, in time t, travels a distance equal to S = u*t + (1/2)*a*t^2. The velocity of the body after time t is equal to v = u + a*t.
When the drop falls, its initial speed is 0. Ignore resistance due to air and take the acceleration due to gravity to be constant and equal to 9.8 m/s^2.
The distance traveled by the drop in the first second is 0*1 + (1/2)*9.8*1 = 4.9 m.
At the start of the 3rd second the speed of the drop is 0 + 9.8*2 = 19.6 m/s. The distance traveled by the drop in the third second is 19.6*1 + (1/2)*9.8*1 = 19.6 + 4.9 = 24.5
The difference between 24.5 and 4.9 is 19.6.
In the 3rd second the drop travels a distance 19.6 m more than it did in the first second after it fell off the ledge.