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

Answer:

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

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

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

$t=\frac{Sf-So}{a}$

We put tha data

$t=\frac{30.47m/s-1.8m/s}{9.8m/s^{2} }$

$t= 2.92s$

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