Courses
Courses for Kids
Free study material
Offline Centres
More

# A hang glider dropped his cell phone from a height of 350 feet. How many seconds did it take for the cell phone to reach the ground?

Last updated date: 29th Feb 2024
Total views: 340.2k
Views today: 9.40k
Verified
340.2k+ views
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:
$1ft=0.305m$
$s=ut+\dfrac{1}{2}a{{t}^{2}}$

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.

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