1. ## 3 short questions

Question 1:
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This one should be simple, I already have the answers to a, b and c. But for some reason I can't seem to remember how to calculate d. That is if the motion of an object (in this case a ball) can be modeled by a parabola, then what was the initial angle the ball was fired at (in radians)? Any pointers on this one would be vastly appreciated.

Question 2:
ImageShack - Hosting :: question42208lu9.jpg

This is how I worked through the problem:

ImageShack - Hosting :: rwer21234oq5.jpg

But obviously my answer is incorrect, so if anyone could help explain where in my steps to solve the question I went wrong, that would be fantastic.

Question 3:

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This question is flooring me at the moment, I have no clue where to start or even what the notation in the second line of the question is referring to. If someone could set me on the path to solving for a, b and c I would be most thankful

2. bump

3. Originally Posted by jpatrie
Question 2:
ImageShack - Hosting :: question42208lu9.jpg

This is how I worked through the problem:

ImageShack - Hosting :: rwer21234oq5.jpg

But obviously my answer is incorrect, so if anyone could help explain where in my steps to solve the question I went wrong, that would be fantastic.
$\displaystyle \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]$

When $\displaystyle i=1$, we see that in the exponent we have $\displaystyle \sum_{l=1}^1 l=1$. Thus, $\displaystyle 1-1=0$

We then end up with the first term in the outermost series: $\displaystyle \left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^0\right)!=1!=1$

Now, when $\displaystyle i=2$:

we see that in the exponent, $\displaystyle \sum_{l=1}^2 j=3$. Thus, $\displaystyle 3-1=2$

Now, let's focus on $\displaystyle \sum_{j=1}^2\prod_{k=1}^j k$

When $\displaystyle j=1$, we have $\displaystyle \prod_{k=1}^1 k=1$

When $\displaystyle j=2$, we have $\displaystyle \prod_{k=1}^2 k=2$

Thus, $\displaystyle \sum_{j=1}^2\prod_{k=1}^j k= 1+2=3$

So we now see that $\displaystyle (3^2)!=9!$

So, we then see that $\displaystyle \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]=1+9!=\dots$

Does this make sense?

--Chris

4. Originally Posted by jpatrie
bump
(1) a) Let the equation of the parapola is

$\displaystyle y = ax^2 + bx + c$

since it passes through (0, 0), put this in eqn, we got, c=0.

So, the eqn becomes, $\displaystyle y = ax^2 + bx$

since it passes through (10, 100),

$\displaystyle 100 = a (10)^2 + b(10)$

$\displaystyle \Rightarrow 100a + 10b = 100$

$\displaystyle \Rightarrow 10a + b = 10 ...................eqn (1)$

Since it passes through (12, 112),

$\displaystyle \Rightarrow 112 = a(12)^2 + b(12)$
$\displaystyle \Rightarrow 144a + 12b = 112$

$\displaystyle \Rightarrow 36a + 3b = 28 .........................eqn (2)$

Now, you solve (1) and (2),to find a and b. we got $\displaystyle a = \frac {-1}{3}, \;\;b= \frac{40}{3}$

Put the value of a and b to get the equation.

equation is $\displaystyle \boxed{\;\;y = \frac{-1}{3}x^2 + \frac{40}{3}x \;\;}$

b) $\displaystyle y = \frac{-1}{3}x^2 + \frac{40}{3}x$

$\displaystyle y = \frac{-1}{3} \left(x^2 - 40x \right)$

$\displaystyle y = \frac{-1}{3} \left(x^2 - 40x + 20^2 - 20^2\right)$

$\displaystyle y = \frac{-1}{3} \left(x - 20 \right)^2 + \frac{400}{3}$

so, the maximum height reached is $\displaystyle \frac{400}{3} \;\;metres$

c) For horizontal distance, put y=0

$\displaystyle 0 = \frac{-1}{3}x^2 + \frac{40}{3}x$

$\displaystyle \Rightarrow x = 0, \;\; 40$

Horizontal distance = 40 -0 = 40 m

Now, you try the last (d) part.

5. Originally Posted by Shyam
(1) a) Let the equation of the parapola is

$\displaystyle y = ax^2 + bx + c$

since it passes through (0, 0), put this in eqn, we got, c=0.

So, the eqn becomes, $\displaystyle y = ax^2 + bx$

since it passes through (10, 100),

$\displaystyle 100 = a (10)^2 + b(10)$

$\displaystyle \Rightarrow 100a + 10b = 100$

$\displaystyle \Rightarrow 10a + b = 10 ...................eqn (1)$

Since it passes through (12, 112),

$\displaystyle \Rightarrow 112 = a(12)^2 + b(12)$
$\displaystyle \Rightarrow 144a + 12b = 112$

$\displaystyle \Rightarrow 36a + 3b = 28 .........................eqn (2)$

Now, you solve (1) and (2),to find a and b. we got $\displaystyle a = \frac {-1}{3}, \;\;b= \frac{40}{3}$

Put the value of a and b to get the equation.

equation is $\displaystyle \boxed{\;\;y = \frac{-1}{3}x^2 + \frac{40}{3}x \;\;}$

b) $\displaystyle y = \frac{-1}{3}x^2 + \frac{40}{3}x$

$\displaystyle y = \frac{-1}{3} \left(x^2 - 40x \right)$

$\displaystyle y = \frac{-1}{3} \left(x^2 - 40x + 20^2 - 20^2\right)$

$\displaystyle y = \frac{-1}{3} \left(x - 20 \right)^2 + \frac{400}{3}$

so, the maximum height reached is $\displaystyle \frac{400}{3} \;\;metres$

c) For horizontal distance, put y=0

$\displaystyle 0 = \frac{-1}{3}x^2 + \frac{40}{3}x$

$\displaystyle \Rightarrow x = 0, \;\; 40$

Horizontal distance = 40 -0 = 40 m

Now, you try the last (d) part.
well, you see part a,b and c are easy, part d was the only part I couldn't figure out, but thanks for confirming my answers for part a,b and c

6. Originally Posted by Chris L T521
$\displaystyle \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]$

When $\displaystyle i=1$, we see that in the exponent we have $\displaystyle \sum_{l=1}^1 l=1$. Thus, $\displaystyle 1-1=0$

We then end up with the first term in the outermost series: $\displaystyle \left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^0\right)!=1!=1$

Now, when $\displaystyle i=2$:

we see that in the exponent, $\displaystyle \sum_{l=1}^2 j=3$. Thus, $\displaystyle 3-1=2$

Now, let's focus on $\displaystyle \sum_{j=1}^2\prod_{k=1}^j k$

When $\displaystyle j=1$, we have $\displaystyle \prod_{k=1}^1 k=1$

When $\displaystyle j=2$, we have $\displaystyle \prod_{k=1}^2 k=2$

Thus, $\displaystyle \sum_{j=1}^2\prod_{k=1}^j k= 1+2=3$

So we now see that $\displaystyle (3^2)!=9!$

So, we then see that $\displaystyle \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]=1+9!=\dots$

Does this make sense?

--Chris
Actually, I believe in the question the exponent of $\displaystyle (\sum_{l=1}^i l)$ is to the power of (-1) NOT subtracting 1

7. Originally Posted by jpatrie
Actually, I believe in the question the exponent of $\displaystyle (\sum_{l=1}^i l)$ is to the power of (-1) NOT subtracting 1
If that was the case, then the second term in the outer most series would be $\displaystyle (\sqrt{3})!$, which is not possible. In order to evaluate $\displaystyle n!$, $\displaystyle n\in\mathbb{Z^+}$, where $\displaystyle \mathbb{Z^+}=\left\{0,1,2,3,...\right\}$. Here, $\displaystyle n$ is irrational.

So I'm pretty sure you are subtracting 1 from that sum in the exponent.

Do you see what I'm trying to say?

--Chris

8. Originally Posted by Chris L T521
If that was the case, then the second term in the outer most series would be $\displaystyle (\sqrt{3})!$, which is not possible. In order to evaluate $\displaystyle n!$, $\displaystyle n\in\mathbb{Z^+}$, where $\displaystyle \mathbb{Z^+}=\left\{0,1,2,3,...\right\}$. Here, $\displaystyle n$ is irrational.

So I'm pretty sure you are subtracting 1 from that sum in the exponent.

Do you see what I'm trying to say?

--Chris
Yea I do understand, I just thought that maybe I had made an error when I found root 3 factorial as the answer.

Thanks

9. Originally Posted by Chris L T521
$\displaystyle \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]$

When $\displaystyle i=1$, we see that in the exponent we have $\displaystyle \sum_{l=1}^1 l=1$. Thus, $\displaystyle 1-1=0$

We then end up with the first term in the outermost series: $\displaystyle \left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^0\right)!=1!=1$

Now, when $\displaystyle i=2$:

we see that in the exponent, $\displaystyle \sum_{l=1}^2 j=3$. Thus, $\displaystyle 3-1=2$

Now, let's focus on $\displaystyle \sum_{j=1}^2\prod_{k=1}^j k$

When $\displaystyle j=1$, we have $\displaystyle \prod_{k=1}^1 k=1$

When $\displaystyle j=2$, we have $\displaystyle \prod_{k=1}^2 k=2$

Thus, $\displaystyle \sum_{j=1}^2\prod_{k=1}^j k= 1+2=3$

So we now see that $\displaystyle (3^2)!=9!$

So, we then see that $\displaystyle \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]=1+9!=\dots$

Does this make sense?

--Chris
Also in general note that $\displaystyle \prod_{k=1}^{j}k=j!$ holds for $\displaystyle j\in\mathbb{N}$ and $\displaystyle \sum\limits_{l=1}^{i}l=\frac{i(i+1)}{2}$ for $\displaystyle i\in\mathbb{N}$.
Hence
$\displaystyle \sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^2\prod_{k=1}^j k\right)^{\left(\sum\limits_{l=1}^i l\right)-1}\right)!\right]$
$\displaystyle =\sum_{i=1}^{2}\left[\left(\left(\sum_{j=1}^{2}j!\right)^{\left(\sum\li mits_{l=1}^i l\right)-1}\right)!\right]$
$\displaystyle =\sum_{i=1}^{2}((1!+2!)^{\frac{i(i+1)}{2}-1})!$
$\displaystyle =\sum_{i=1}^{2}(3^{\frac{i(i+1)}{2}-1})!$
$\displaystyle =1!+9!=362881$
This pathern works when $\displaystyle 2$ is replaced by any positive integer $\displaystyle n$.