Question

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?

Answer 1

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

Step-by-step 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

`Sf^2= So^2 + (2*a*x)`

Where Sf= final speed

We put the data

`Sf^2= 928.36m^(2)/s^(2)`

We need to calculate the square root

`Sf= 30.47m/s`

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

`t=(Sf-So)/(a)`

We put tha data

`t=(30.47m/s-1.8m/s)/(9.8m/s^(2) )`

`t= 2.92s`

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