Question

Gravity: A hang glider dropped his cell phone from a height of 350 feet. Use the formula$\mathit{t}\mathbf{=}\frac{\sqrt{\mathbf{h}}}{\mathbf{4}}$ to find how many seconds it took for the cell phone to reach the ground.

Short answer

It will take 4.7seconds for the cell phone to hit the ground.

Step-by-step solution

Step 1. Given Information:

A cell phone have been dropped from the height of 350 feet above the ground,

Formula : $\mathit{t}\mathbf{=}\frac{\sqrt{\mathbf{h}}}{\mathbf{4}}$

Step 2. Assign the value to the equation:

Let t be the no of seconds in the formula,

h is the given height,

Substitute in the given information in the formula :

$$\begin{gathered} \mathit{t}\mathbf{=}\frac{\sqrt{\mathbf{350}}}{\mathbf{4}} \\ \mathit{t}\mathbf{=}\frac{\sqrt{{\mathbf{\left(}\mathbf{25}\mathbf{\right)}}^{\cancel{\mathbf{2}}}\mathbf{\times }\mathbf{14}}}{\mathbf{4}} \\ \mathit{t}\mathbf{=}\frac{\mathbf{5}\mathbf{\times }\sqrt{\mathbf{14}}}{\mathbf{4}}\mathbf{=}\frac{\mathbf{5}\mathbf{\times }\mathbf{3}\mathbf{.}\mathbf{74}}{\mathbf{4}}\mathbf{=}\frac{\mathbf{18}\mathbf{.}\mathbf{7}}{\mathbf{4}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{675} \\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{7}\mathbf{}\mathit{s}\mathit{e}\mathit{c}\mathit{o}\mathit{n}\mathit{d}\mathit{s}\mathbf{(}\mathit{R}\mathit{o}\mathit{u}\mathit{n}\mathit{d}\mathbf{}\mathit{a}\mathit{p}\mathit{p}\mathit{r}\mathit{o}\mathit{x}\mathit{i}\mathit{m}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathit{o}\mathit{n}\mathit{e}\mathbf{}\mathit{d}\mathit{e}\mathit{c}\mathit{i}\mathit{m}\mathit{a}\mathit{l}\mathbf{}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}\mathbf{)} \end{gathered}$$

Step 3. Checking the solution: 

Here time t = 4.7,

$$\begin{gathered} \mathit{t}\mathbf{=}\frac{\sqrt{\mathbf{350}}}{\mathbf{4}} \\ \mathbf{4}\mathbf{.}\mathbf{7}\mathbf{=}\frac{\sqrt{\mathbf{350}}}{\mathbf{4}} \\ \mathbf{4}\mathbf{.}\mathbf{7}\mathbf{=}\frac{\sqrt{{\mathbf{\left(}\mathbf{25}\mathbf{\right)}}^{\cancel{\mathbf{2}}}\mathbf{\times }\mathbf{14}}}{\mathbf{4}} \\ \mathbf{4}\mathbf{.}\mathbf{7}\mathbf{=}\frac{\mathbf{5}\mathbf{\times }\sqrt{\mathbf{14}}}{\mathbf{4}}\mathbf{=}\frac{\mathbf{5}\mathbf{\times }\mathbf{3}\mathbf{.}\mathbf{74}}{\mathbf{4}}\mathbf{=}\frac{\mathbf{18}\mathbf{.}\mathbf{7}}{\mathbf{4}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{675} \\ \mathbf{4}\mathbf{.}\mathbf{7}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{7}\mathbf{}\mathit{s}\mathit{e}\mathit{c}\mathit{o}\mathit{n}\mathit{d}\mathit{s} \\ \mathit{H}\mathit{e}\mathit{n}\mathit{c}\mathit{e}\mathbf{}\mathit{P}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{d}\mathbf{.} \end{gathered}$$