MHB 3-42 Where on ground (relative to position of the helicopter

AI Thread Summary
The discussion focuses on calculating the landing position of a stunt woman dropping from a helicopter 30 meters above the ground, moving upward at 10 m/s and horizontally at 15 m/s. The equations of motion are applied to determine the time of fall, which is approximately 3.7 seconds, and the horizontal distance traveled during this time. The horizontal displacement is calculated to be about 55.5 meters due south from the drop position. The calculations ignore air resistance, simplifying the physics involved. The final conclusion indicates that the foam mats should be placed 55.5 meters south of the helicopter's position at the time of the drop.
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ok this is out of an old textbook and possibly already posted here so..

A movie stunt woman drops from a helicopter that is 30.0 m above the ground and moving with a constant velocity whose components are 10.0 m/s upward and 15.0 m/s horizontal and toward the south. Ignore air resistance.
[a.] Where on the ground (relative to the position of the helicopter when she drops) should the stunt woman have placed the foam mats that break her fall?

well so far... but not sure
$y=y_0+V_0 yt+.5A yt^2$
$0=30+10t[-.5(9.8)t^2]$
so $0=-4.9t^2+10t+30$

later [b.] Draw x-t, y-t, v ft, and v y-t graphs of the motion
 
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$\Delta x = v_{x_0} \cdot t$

$\Delta y = v_{y_0} \cdot t - \dfrac{1}{2}gt^2$

solve the quadratic for $t$, then calculate $\Delta x$
 
ok not sure why this doesn't render but it my understanding of the problem
\begin{tikzpicture}[xscale=.4,yscale=.2]
\draw [very thick] (-1,0) -- (20,0);
\draw [thin] (0,0) -- (0,30);
\draw [dashed][->] (0,30) -- (15,30);
\draw [dashed][->] (0,30) -- (0,40);
\draw [dashed][->] (0,30) -- (15,40);
\node
at (0,30) {30 m};
\node
at (15,30) {15 m/s};
\node [above] at (0,40) {10 m/s};
\end{tikzpicture}
at (15,30) {15 m/s};
\node [above] at (0,40) {10 m/s};
\end{tikzpicture}​
 
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$\dfrac{1}{2}gt^2 - v_{y_0} \cdot t + \Delta y = 0 \implies t = \dfrac{v_{y_0} + \sqrt{(v_{y_0})^2 - 2g\Delta y}}{g}$

for the given values, $t \approx 3.7 \, sec$

$\Delta x = v_{x_0} \cdot t \approx 55.5 \, m$ due South of the drop position.
 
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