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Given:
The mass of the phone is,
[tex]m=0.264\text{ kg}[/tex]The initial height of the phone is,
[tex]H=13\text{ m}[/tex]The initial potential energy of the phone will be converted into kinetic energy when it hits the ground.
If the velocity of the phone is v at the time of touching the ground we can write,
[tex]\begin{gathered} \frac{1}{2}mv^2=mgH \\ v=\sqrt[]{2gH} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} v=\sqrt[]{2\times9.81\times13} \\ =15.9\text{ m/s} \end{gathered}[/tex]Hence, the phone will have a speed of 15.9 m/s.