A stone is thrown straight down from the edge of a roof, 725 feet above the ground, at a speed of 7 feet per second. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 3 seconds later? At what time does the stone hit the ground? What is the velocity of the stone when it hits the ground?

Guest Jan 29, 2015

#1**+8 **

1. Use s = ut + (1/2)at^{2}, where s = distance travelled, u = initial velocity, a = acceleration, t = time.

s = 7*3 +(1/2)*32*9 = 165 ft., so the stone is 725 - 165 = 560 ft off the ground.

2. The stone hits the ground when s = 725, so: 725 = 7*t +(1/2)*32*t^{2}

Solving this quadratic we get

$${\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{725}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5\,161}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}\right)}{{\mathtt{32}}}}\\

{\mathtt{t}} = {\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5\,161}}}}{\mathtt{\,-\,}}{\mathtt{7}}\right)}{{\mathtt{32}}}}\\

\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = -{\mathtt{6.953\: \!759\: \!395\: \!873\: \!178\: \!7}}\\

{\mathtt{t}} = {\mathtt{6.516\: \!259\: \!395\: \!873\: \!178\: \!7}}\\

\end{array} \right\}$$

The only positive solution is t = 6.516 seconds

3. Use v^{2} = u^{2} + 2as where v is final velocity.

v^{2} = 49 + 2*32*725

$${\mathtt{v}} = {\sqrt{{\mathtt{49}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{32}}{\mathtt{\,\times\,}}{\mathtt{725}}}} \Rightarrow {\mathtt{v}} = {\mathtt{215.520\: \!300\: \!667\: \!941\: \!719\: \!5}}$$

v = 215.52 ft/sec

.

Alan
Jan 29, 2015

#1**+8 **

Best Answer

1. Use s = ut + (1/2)at^{2}, where s = distance travelled, u = initial velocity, a = acceleration, t = time.

s = 7*3 +(1/2)*32*9 = 165 ft., so the stone is 725 - 165 = 560 ft off the ground.

2. The stone hits the ground when s = 725, so: 725 = 7*t +(1/2)*32*t^{2}

Solving this quadratic we get

$${\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{725}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5\,161}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}\right)}{{\mathtt{32}}}}\\

{\mathtt{t}} = {\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5\,161}}}}{\mathtt{\,-\,}}{\mathtt{7}}\right)}{{\mathtt{32}}}}\\

\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = -{\mathtt{6.953\: \!759\: \!395\: \!873\: \!178\: \!7}}\\

{\mathtt{t}} = {\mathtt{6.516\: \!259\: \!395\: \!873\: \!178\: \!7}}\\

\end{array} \right\}$$

The only positive solution is t = 6.516 seconds

3. Use v^{2} = u^{2} + 2as where v is final velocity.

v^{2} = 49 + 2*32*725

$${\mathtt{v}} = {\sqrt{{\mathtt{49}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{32}}{\mathtt{\,\times\,}}{\mathtt{725}}}} \Rightarrow {\mathtt{v}} = {\mathtt{215.520\: \!300\: \!667\: \!941\: \!719\: \!5}}$$

v = 215.52 ft/sec

.

Alan
Jan 29, 2015