Author Topic: Am i going over time? (Read 4090 times) Tweet Share

dejan91 · « **Reply #15 on:** October 25, 2009, 10:28:19 pm »

You can find the cut-off's on the vcaa website: http://www.vcaa.vic.edu.au/vce/statistics/2007/statssect3.html#H3N10378

And with that question you have to use related rates of change: $\frac{dh}{dt} = \frac{d\theta}{dt} * \frac{dh}{d\theta}$ using $tan(\theta) = \frac{h}{500}$ .

aronno · « **Reply #16 on:** October 25, 2009, 10:30:07 pm »

say what? not the Q.4 from the multi's i meant the one from the short answer

dejan91 · « **Reply #17 on:** October 25, 2009, 10:31:46 pm »

Lol I saw "launching" and remembered something with a rocket in the '07 one. Woops

philza · « **Reply #18 on:** October 25, 2009, 10:36:35 pm »

:S i'm spendiong 40-50 aswell T.T

GerrySly · « **Reply #19 on:** October 25, 2009, 10:42:45 pm »

Quote from: aronno on October 25, 2009, 10:13:31 pm

hey guys, i am specifically having trouble with working out the landing angles or launching angle questions. Similar to that of vcaa 07 Q.4 (c)
please help.

I believe you dot product the velocity vector with a vector on the ground (get rid of the k component)

$\begin{align*} \tilde{h}&=30\tilde{i}-40\tilde{j}\\ \tilde{v}&=30\tilde{i}-40\tilde{j}-4\tidle{k}\\ \tilde{h}\cdot \tilde{v}&=|\tilde{h}||\tilde{v}|\cos{(\theta)}\\ 2500&=\sqrt{30^2+40^2}\cdot \sqrt{30^2+40^2+4^2}\cos{(\theta)}\\ \cos{(\theta)}&=\frac{2500}{\sqrt{2500}\cdot\sqrt{2516}}\\ &=\frac{25}{\sqrt{629}}\\ \implies \theta &= 4.57^o \end{align*}$

aronno · « **Reply #20 on:** October 25, 2009, 11:17:05 pm »

thanks BRO!!!
btw i've seen questions use tan(theta) = [(a)^2]/{((b)^2+(c)^2))^(1/2)}
to determine the angle. what does this relate to? or what is the significance

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Author Topic: Am i going over time? (Read 4090 times) Tweet Share

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