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 #1
avatar+2441 
+2

Heron's method of finding the square root of a number is interesting, but this is what you do

 

1. Take a reasonable guess of what the answer would be (to the integer)

2. Divide the radicand by your guess.

3. Find the average of your guess and the radicand divided by your guess.

4. Repeat

 

Ok, I will show how to estimate the values above \(\sqrt{120}\).

 

1. Take a Reasonable Guess

 

The first step to estimating the value of \(\sqrt{120}\) is to take a guess. I would suggest leaving it as an integer.

 

I happen to know that \(11^2=121\), so 11 should be a very good guess to what the actual decimal expansion is of \(\sqrt{120}\) is. You may overestimate or underestimate; it should not make a difference. However, the closer your original guess is, the more accurate your result will be.

 

2. Divide the Radicand by Your Guess

 

The radicand, if you are unaware, is the number inside of the radical. For example, in this case, the radicand of \(\sqrt{120}\) is 120. The radicand of \(\sqrt{2}\) is 2. We want to take 120 and divide it by our guess, 11.

 

\(\frac{120}{11}=10.\overline{9090}\), so I will leave 120/11 as it is because the fraction cannot be simplified any further.

 

3. Find the Average of Your Guess and the Radicand Divided by Your Guess

 

In this step, simply take average of your guess, 11, and the radicand divided by your guess, which was calculated to be \(\frac{120}{11}\). Let's do that:

 

\(\frac{11+\frac{120}{11}}{2}\) Now, just simplify. I will convert 11 into an improper fraction. \(11*\frac{11}{11}=\frac{121}{11}\)
\(\frac{\frac{121}{11}+\frac{120}{11}}{2}\) Find the sum of the numerator by adding the fractions.
\(\frac{\frac{241}{11}}{2}=\frac{241}{11*2}=\frac{241}{22}\) Unfortunately, 241/11 cannot be simplified further. Now, I will utilize a fraction rule that states that \(\frac{\frac{a}{b}}{c}=\frac{a}{b*c}\)
   

 

You can continue this process to get a more accurate decimal approximation, too, but this fraction is OK for just an approximation. Here's how close this approximation was:

 

\(\frac{241}{22}=10.95\overline{45}\)

\(\sqrt{120}\approx10.95445\)

 

That's really accurate, wouldn't you agree?

 

Ok, I will do the next one, too:

 

1. Take a Reasonable Guess

 

Yet again, we must make a guess for the \(\sqrt{85}\).

 

I happen to know that \(9^2=81\), so this seems like a good estimate. 

 

2. Divide the Radicand by Your Guess

 

\(\frac{85}{9}\) cannot be simplified further because the numerator and denominator are co-prime, so we will have to leave it as is. Also, \(\frac{85}{9}=9.\overline{4}\), so it would be difficult to calculate it in decimal form.

 

3. Find the Average of Your Guess and the Radicand Divided by Your Guess

 

Just like the previous problem, we are finding the average of 81 and \(\frac{85}{9}\):

 

\(\frac{9+\frac{85}{9}}{2}\) First, convert 9 into an improper fraction such that the denominator is 9.
\(\frac{\frac{81}{9}+\frac{85}{9}}{2}\) Simplify the numerator.
\(\frac{\frac{166}{9}}{2}\) Just like the previous problem, we will utilize the fraction rule \(\frac{\frac{a}{b}}{c}=\frac{a}{b*c}\)
\(\frac{\frac{166}{9}}{2}=\frac{166}{2*9}=\frac{166}{18}\) In this case, the numbers 166 and 18 have a GCF of 2, so we can simplify this faction further.
\(\frac{166}{18}\div\frac{2}{2}=\frac{83}{9}\)  
   

 

Let's see how close our approximation is:

 

\(\frac{83}{9}=9.\overline{2}\)

\(\sqrt{85}\approx9.2195\)

 

I would say that that is a good approximation.

 

--------------------------------------------------------------------------------

 

And finally 20.

 

My guess will be 4 as \(4^2=16\), so that seems like a good approximation.

 

\(\frac{20}{4}=5\). Now that I have divided the radicand by my guess. We can do the next step:

 

\(\frac{4+5}{2}\) Find the sum in the numerator.
\(\frac{9}{2}\)  
   

 

As a check, let's see how close our approximation truly is:

 

\(\frac{9}{2}=4.5 \)

\(\sqrt{20}\approx4.47214\)

 

That seems pretty good! 

 #5
avatar+2440 
+4

Solutions for A - E

 

\(\text{Position and Time }\\ \begin{array}{|rcll|} \hline a_1) \text { position }\\ h &=& v_i(t) -\dfrac{1}{2} g(t)^2 \\ h &=& 15t - 4.9t^2 \\ t\tiny \; \text{ @} \small \text { 1 and 4}\\ h &=& 15(1) - 4.9(1)^2 &=&10.1m \\ h &=& 15(4) - 4.9(4)^2 &=&-18.4m \\ a_2) \text { velocity} \\ v_f &=& v_i – gt \\ v_f &=& 15 - 9.8t \\ \text{t @ 1 and 4} \\ v_f &=& 15 - 9.8(1) &=&5.20m/s \\ v_f &=& 15 - 9.8(4) &=& -24.2m/s \\ \hline \end{array}\\ \text { at 1 second: ball is 10.1m (above) and moving @ 5.20 m/s (up)} \\ \text { at 4 seconds: ball is -18.4m (below) and moving -24.2 m/s(down)} \\\)

 

\(b)\text {Gravity is slowing ball} \\ \begin{array}{|rcll|} \hline v^2 &=&v_0^2 +2(a)(\Delta y) \\ v^2 &=&(15)^2 +2(-9.8)(5m) \leftarrow \text{ use negative acceleration} \\ v &=&\pm \sqrt{127} \text{ (use positive result) } \\ v &=& 11.27m/s \\ \hline \end{array}\)

 

\(c)\text {Maximum height occurs when the vertical velocity is zero (0)} \\ \begin{array}{|rcll|} \hline a &=& \dfrac {(v_f – v_i)}{t} \\ t &=& \dfrac {(v_f – v_i)}{a} \\ t &=& \dfrac {(0 - 15)}{(- 9.8)} \\ t &=& \dfrac{-15}{-9.8} \\ t &=& 1.53s \\ \hline \end{array}\)

 

\(d) \text{ Acceleration is constant (g)} \\\)

 

\(e) \text{ Time after displacement }\\ \begin{array}{|rcll|} \hline t &=& \dfrac {(v_f – v_i)}{a} \\ v_f^2 &=& v_i^2 + 2(a)(\Delta y)\\ v_f &=& \pm \ sqrt{(v_i^2 + 2(a)(\Delta y)} \leftarrow \small \text {Use negative result; the ball is moving downward } \\ t &=& \dfrac{-\sqrt{(v_i^2 + 2(a)(\Delta y)} + (v_i)}{(-g)} \\ t &=& \dfrac{-\sqrt{(15^2 + 19.6*5)} + 15)}{-9.8}\\ t &=& 3.36s \\ \hline \end{array}\)

 

 

For related effects of gravity, see this.

Aug 1, 2017
 #3
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0
Aug 1, 2017
Jul 31, 2017
 #2
avatar+118608 
+1
Jul 31, 2017
 #7
avatar+118608 
0
Jul 31, 2017
 #1
avatar+14923 
+1

Hello Guest!

 

You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at the point even with the roof railing with an upward speed of 15.0 m/s, the ball is then in free fall. On its way back down, it just misses the railing. At the location of the building, g = 9.8m/s^2.

 

Find :

(a) the position and the velocity of the ball 1.0s and 4.0s after leaving your hand.

(b) The velocity when the ball is 5.0m above the railing.

(c) The maximum height reached and the time at which it is reached. 

(d) The acceleration of the ball when it is at its maximum height.

(e) Find the time when the ball is 5.0m below the roof railing.

 

(a) the position and the velocity of the ball 1.0s and 4.0s after leaving your hand.

 

\(h=vt-\frac{1}{2}gt^2\\ h_{1.0}=\frac{15m}{s}\cdot 1s-\frac{1}{2}\cdot\frac{9.8m}{s^2}\cdot 1^2s^2\\ \color{blue}h_{1.0}=10.1m\ over\ the\ railing\)

 

\(h_{4.0}=\frac{15m}{s}\cdot 4s-\frac{1}{2}\cdot\frac{9.8m}{s^2}\cdot 4^2s^2\\ h_{4.0}=60m-79.2m=-19.2m\\ \color{blue}h_{4.0}=19.2m\ under\ the\ railing\)

 

\(v=v_0-gt\\ v_{1.0}=\frac{15m}{s}-\frac{9.8m\cdot1s}{s^2}=5.2\frac{m}{s}\\ \color{blue}v_{1.0}=5.2\frac{m}{s}\ up \)

 

\(v_{4.0}=\frac{15m}{s}-\frac{9.8m\cdot4s}{s^2}=-24.2\frac{m}{s}\\ \color{blue}v_{4.0}=24.2\frac{m}{s}\ down\)

 

(b) The velocity when the ball is 5.0m above the railing.

 

\(v=v_0-\sqrt{2gh}\\ v_{(+5m)}=\frac{15m}{s}-\sqrt{\frac{2\cdot 9.8m\cdot 5m}{s^2}}=5.1\frac{m}{s}\\ \color{blue}v_{(+5m)}=5.1\frac{m}{s}\ up\)

 

 

(c) The maximum height reached and the time at which it is reached

 

\(E=\frac{mv^2}{2}=mgh\\ h=\frac{mv^2}{2mg}=\frac{v^2}{2g}\)

 

\(h_{max}=\frac{15^2m^2s^2}{2\cdot 9.8ms^2}\\ \color{blue} h_{max}=11.480m\ over\ the\ railing. \)

 

\(v=gt\\ t_{rise}=\frac{v}{g}=\frac{15ms^2}{9.8ms}\\ \color{blue} t_{rise}=1.531s\)

 

(d) The acceleration of the ball when it is at its maximum height.

 

The ball is weightless throughout the journey.
For the ball, the acceleration is zero everywhere. It is not exposed to any force.
For the viewer the acceleration is at the highest point and in the case g = 9.8m / s²

 

(e) Find the time when the ball is 5.0m below the roof railing.

 

\(s=vt-\frac{1}{2}gt^2\\ \frac{1}{2}gt^2-vt+s=0\\ t^2-\frac{2v}{g}t+\frac{2s}{g}=0\)

 \(t=-\frac{p}{2}+\sqrt{(\frac{p}{2})^2-q}\\ \color{blue} t=\frac{v}{g}+\sqrt{\frac{v^2}{g^2}-\frac{2s}{g}}\)    

\(t=\frac{15m/s}{9.8m/s^2}+\sqrt{(\frac{15}{9.8}s)^2-\frac{-2\cdot 5m}{9.8m/s^2}}\\ t=1.531s+\sqrt{2.343s^2+1,020s^2}\\ t=1.531s+1.834s\\ \color{blue}t=3.365s\ (5m\ under\ the\ railing)\)

 

laugh  !

Jul 31, 2017

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