Answer

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**Hint:**A cell phone falls from a certain height above the ground therefore, it is under the influence of gravity. Therefore we can use equations of motion to solve the problem. Substituting corresponding values in the equation, the time can be calculated. Convert the equations as required.

**Formula used:**

$1ft=0.305m$

$s=ut+\dfrac{1}{2}a{{t}^{2}}$

**Complete answer:**

Given, a cell phone falls freely from a height of 350 feet.

We know that,

$1ft=0.305m$

Using the above relation we convert 350 feet into metres as-

$\begin{align}

& 350ft=350\times 0.305m \\

& \Rightarrow 350ft=106.75m \\

\end{align}$

The cell phone is falling freely, therefore its acceleration will be $10m{{s}^{-2}}$.

Using the following equation of motion, we get,

$s=ut+\dfrac{1}{2}a{{t}^{2}}$

Here, $s$ is the distance travelled

$u$ is the initial velocity

$t$ is the time taken

$a$ is the acceleration of the body

Substituting given values in the above equation, we get,

$\begin{align}

& 106.75=0+\dfrac{1}{2}\times 10{{t}^{2}} \\

& \Rightarrow \dfrac{106.75\times 2}{10}={{t}^{2}} \\

& \Rightarrow 21.35={{t}^{2}} \\

& \Rightarrow t=4.62s \\

& \therefore t\approx 5s \\

\end{align}$

The cell phone reached the ground in about $5s$.

Therefore, it took about $5s$ for the cell phone to reach the ground.

**Additional Information:**

The equations of motions are used to describe the motion of an object in a straight line when the acceleration is constant. It gives us the relation between displacement, initial velocity, final velocity, acceleration and time taken. Some equations of motion are; ${{v}^{2}}={{u}^{2}}+2as$, $v=u+at$ and $s=ut+\dfrac{1}{2}a{{t}^{2}}$.

**Note:**

According to Newton’s second law, when a body undergoes acceleration, an external force is acting on it. The acceleration of the cell phone is constant therefore; equations of motions can be applied. In freely falling condition, the body is under acceleration due to gravity. The potential energy of the cell phone will be highest at the top which will then convert into kinetic energy throughout the motion.

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