Question

A stone was dropped off a cliff and hit the ground with a speed of 120 ft/s. What is the height of the cliff?

Short answer

The height of the cliff is 225 ft.

Step-by-step solution

Explanation

Here we use acceleration due to gravity\[ - 32\frac{{ft}}{s}\].

Now we know that acceleration function.

\[at = v'\left( t \right)\_\_\_\_\_(1)\].

Here \[v'\left( t \right)\]is the first derivative of the velocity function\[v\left( t \right)\].

Expression for velocity function\[v\left( t \right)\].

\[v\left( t \right) = s'\left( t \right)\_\_\_\_\_\_(2)\].

Here \[s'\left( t \right)\]is the first derivative of the displacement function s(t).

Find the height of the cliff

Now acceleration due to gravity is\( - 32\frac{{ft}}{s}\).

Now \(a = - 32\).

We written derivative of\(a\),

\(a' = - 32t + c\).

When it dropped \(t = 0\)and \(v = 0\)

\(\begin{aligned}{c}v &= - 32\left( 0 \right) + c\\0 &= 0 + c\\c &= 0\\v &= - 32t\end{aligned}\)

Given in our question \(v = 120\).

\(\begin{aligned}{c}120 &= - 32t\\t &= - \frac{{120}}{{32}}\\ &= - \frac{{15}}{4}\end{aligned}\)

Here we have written derivative of \(v\).

\(v'' = - 32\frac{{{t^2}}}{2} + k\)

We know that\(s = v''\), we explain it in the above.

\(\begin{aligned}{c}s &= - 32\frac{{{t^2}}}{2} + k\\ &= - 16{t^2} + k\end{aligned}\)

When it hit the ground, \(t = - \frac{{15}}{4}\), and \(s = 0\), then,

\(\begin{aligned}{c}0 &= - 16{\left( {\frac{{15}}{4}} \right)^2} + k\\ - k &= - 16\left( {\frac{{225}}{{16}}} \right)\\k &= 225\end{aligned}\)

We get,

\(s = - 16{t^2} + 225\)

The height of the cliff is 255 ft.