The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
[tex]g=9.81 m/s^2[/tex]
So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
[tex]y(t) = \frac{1}{2}gt^2 [/tex]
And if we substitute t=2.7 s, we find the distance covered:
[tex]y(t)= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2=35.8 m [/tex]
