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

[karush]

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

 

[skeeter]

$\Delta x = v_{x_0} \cdot t$

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

solve the quadratic for $t$, then calculate $\Delta x$

 

[karush]

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}​

 

[skeeter]

$\dfrac{1}{2}gt^2 - v_{y_0} \cdot t + \Delta y = 0 \implies t = \dfrac{v_{y_0} + \sqrt{(v_{y_0})^2 - 2g\Delta y}}{g}$

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.