An inquisitive physics student and mountian climber climbs a 47.2 m cliff that overhangs a calm pool of water. he throws two stones vertically downward, 0.8 s apart and observes that they cause a single splash. the first stone has an initial velocity of 1.8 m/s. how long after release of the first stone do the two stones hit the water?

The two stoned hit the water after 2.92s after the first stone satat to drop

Explanation:

We have this data:

x = distance = 47.2m

So = initial speed =  1.8m/s

A = In this case is equal to gravity = 9.8 m/s2

First we are going to calculate the final speed, We are going to use this equation

Where Sf= final speed

We put the data

We need to calculate the square root

The final speed is 30.47m/s

The next steep is calculate the time that the first stone spend during its drop.

We are going to use the next equation

We put tha data

We do not use 0.8s because we calculate all the things with the first stone, This stone spend more time that the second stone

1/0.14nm = 10^10/1.4 a lot

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