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June 12th, 2018, 05:48 PM   #1
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Exclamation Given a velocity vector and a point, find the angle (urgent)

Hello! So I am doing this vectors as velocity question and I need help solving it. Basically, James Bond is driving 70.9km/h [N] and launches himself off a cliff. A helicopter is waiting for him 20m North and 15m West from him (at the same elevation). If a large gust of wind travelling 54km/h [W] hits him as he jumps from the car, what angle does he need to jump out of the car from in order to land in the helicopter?

So what I did so far was, I found the velocity vector that he is travelling if he jumps directly north from the car and that is:

which gives you 89.12km/h [N37.3W]
I do not know how to figure the angle he should jump at to get to the point.

There isn't a vectors category and since in Canada the course is called "calculus and vectors" I just posted it here; sorry if it isn't supposed to be here.

Last edited by skipjack; June 13th, 2018 at 12:40 AM.
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July 19th, 2018, 03:58 AM   #2
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The problem says "a large gust of wind travelling 54 km/h [W] hits him as he jumps from the car". Are we to assume that this velocity vector will be added to his velocity? Wind does not work that way! If you were talking about a boat traveling on a river, then the boat is carried on the river so its velocity is added to the boat's. James Bond is not being "carried" by the wind. In order to answer the question, as asked, we would need to know a "coefficient of friction" between Bond and the wind.

Ignoring that error, take Bond's initial jump to make angle $\displaystyle \theta$ with north, angle $\displaystyle \phi$ with the horizontal plane. Then his initial velocity vector is $\displaystyle \left<v \sin(\theta), v \cos(\theta), v \cos(\phi)\right>$. Since the only force on him is -mg is the vertical direction (again, since we are not given a coefficient of friction we must assume the wind adds it velocity to his, not affecting his acceleration) we have, as his acceleration vector, $\displaystyle \vec{a}= \left<0, 0, -g\right>$. Integrating his velocity vector would be $\displaystyle \vec{v}= \left<C_1, C_2, -gt+ C_3\right>$. However, we are told that the wind (which, against all physical laws we are apparently supposed to assume adds itself to his velocity) is [math]\left< 0, 54, 0\right> so we must have [math]\vec{v}= \left< C_1, C_2+ 54, -gt+ C_3\right> In order to determine the constants, we use his initial velocity vector: $\displaystyle \left<C_1, C_2, C_3\right>= \left<70.9 \sin(\theta), 70.9 \cos(\theta)+ 54, 70.9 \cos(\phi)\right>$.

Integrating again, his position vector is given by $\displaystyle \vec{x(t)}= \left<70.9 \sin(\theta)t+ D_1, 70.9 \cos(\theta)t+ 54t+ D_2, -\frac{g}{2}t^2+ 70.9 \cos(\phi)t+ D_3\right>$. Taking his initial position at the point where he launches the car to be (0, 0, 0), we have $\displaystyle \vec{v(0)}= \left<D_1, D_2, D_3\right>= <0, 0, 0>$ so his position, at time t, is given by $\displaystyle \vec{x(t)}= \left<70.9 \sin(\theta)t, 70.9 \cos(\theta)+ 54t , -\frac{g}{2}t^2+ v \cos(\phi)t\right>$.

Now, the helicopter is positioned at " 20m North and 15m West from him (at the same elevation)" or at $\displaystyle \left< 20, -15, 0\right>$ so that we must have, for some t, $\displaystyle \left<70.9 \sin(\theta)t, 70.9 \cos(\theta)+ 54t, -\frac{g}{2}t^2+ 70.9 \cos(\phi)t\right>= \left< 20, -15, 0\right>$.

Solve the three equations $\displaystyle 70.9 \sin(\theta)t= 20$, $\displaystyle 70.9 \cos(\theta)+ 54t= -15$, $\displaystyle -\frac{g}{2}t^2+ 70.9 \cos(\phi)t= 0$ for $\displaystyle \theta$, $\displaystyle \phi$, and $\displaystyle t$.

Last edited by skipjack; July 19th, 2018 at 07:26 AM.
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