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# 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 ac

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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?

physics
Guest Jan 29, 2015

#1
+26750
+8

1.  Use s = ut + (1/2)at2, 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*t2

$${\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{725}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5\,161}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}\right)}{{\mathtt{32}}}}\\ {\mathtt{t}} = {\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5\,161}}}}{\mathtt{\,-\,}}{\mathtt{7}}\right)}{{\mathtt{32}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = -{\mathtt{6.953\: \!759\: \!395\: \!873\: \!178\: \!7}}\\ {\mathtt{t}} = {\mathtt{6.516\: \!259\: \!395\: \!873\: \!178\: \!7}}\\ \end{array} \right\}$$

The only positive solution is t = 6.516 seconds

3. Use v2 = u2 + 2as  where v is final velocity.

v2 = 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

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Alan  Jan 29, 2015
#1
+26750
+8

1.  Use s = ut + (1/2)at2, 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*t2

$${\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{725}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5\,161}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}\right)}{{\mathtt{32}}}}\\ {\mathtt{t}} = {\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5\,161}}}}{\mathtt{\,-\,}}{\mathtt{7}}\right)}{{\mathtt{32}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = -{\mathtt{6.953\: \!759\: \!395\: \!873\: \!178\: \!7}}\\ {\mathtt{t}} = {\mathtt{6.516\: \!259\: \!395\: \!873\: \!178\: \!7}}\\ \end{array} \right\}$$

The only positive solution is t = 6.516 seconds

3. Use v2 = u2 + 2as  where v is final velocity.

v2 = 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

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Alan  Jan 29, 2015
#2
+19620
+5

3. also final velocity v:

$$v=v_0+at_{\small{\text{stone hits the ground}}} \quad v_0 = 7\quad t = 6.51625939587s \\ v = 7 + 32 * 6.51625939587 = 215.5203 \ \frac{ft}{s}\\$$

heureka  Jan 29, 2015
#3
+87294
+5

Here's a graph of the position function and the velocity fuction that demonstrates Alan's and Heureka's answers....

GRAPH

CPhill  Jan 29, 2015