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In a fit of jealous rage, your significant other tosses your phone off a cliff. Determine how fast your phone with a mass of 0.264 kg is moving as it hits the ground 13 m below. (2 decimal places)

Sagot :

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.

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