Question

1. a stone falls from the edge of a cliff for 6.7 meters and hits still water. how fast is it moving just as it hits the water (at its 6.7 - meter displacement)?

Explanation:

Step1: Identify the relevant kinematic - equation

The kinematic equation $v^{2}=v_{0}^{2}+2ax$ is used. Here, the initial velocity $v_{0} = 0$ m/s (starts from rest), the acceleration $a = g=9.8$ m/s² (acceleration due to gravity near the Earth's surface), and the displacement $x = 6.7$ m.

Step2: Substitute the values into the equation

Since $v_{0}=0$, the equation simplifies to $v^{2}=2ax$. Substituting $a = 9.8$ m/s² and $x = 6.7$ m, we get $v^{2}=2\times9.8\times6.7$.
$v^{2}=131.32$.

Step3: Solve for $v$

Take the square - root of both sides: $v=\sqrt{131.32}$.
$v\approx11.46$ m/s.

Answer:

$11.46$ m/s