Question

A plant falls from a window ledge 3.5 m above the sidewalk. How fast is the plant moving when it hits the sidewalk?

Answer 1

The motion of the plant is a free-fall motion, with acceleration a=g=9.81 m/s^2, so we can use the following relationship:
$2aS = v_f^2 -v_i^2$
where S=3.5 m is the distance covered by the plant from the window to the sidewalk, vf is the final speed and vi=0 is the initial speed of the plant.

Substituting numbers, we can find the value of the speed when the plant hits the ground:
$v_f = \sqrt{2aS} = \sqrt{2 (9.81 m/s^2)(3.5 m)}=8.3 m/s$