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November 4th, 2019, 08:08 PM   #1
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Question How can I find the speed of a falling object as seen from a car while in motion?

The problem is as follows:
A car driving in a city has a constant velocity of $15\hat{i}-15\hat{j}\frac{m}{s}$ with respect to the ground in the highway. However at $t=0\,s$ the driver in the car sees a lady in highest floor of a nearby building dropping a bucket which was at rest. Find the instant (in seconds) from which the driver situated in the car will measure the speed of the bucket to be $-15\,i\frac{m}{s}$. You may consider $\vec{g}=-10\,\hat{j}\frac{m}{s^{2}}$
The alternatives given on my book are as follows:

$\begin{array}{ll}
1.&0.5\,s\\
2.&1\,s\\
3.&1.5\,s\\
4.&2\,s\\
5.&2.5\,s\\
\end{array}$

How am I supposed to relate the quantity of the velocity and the instant?.

The only equation which I can recall for the position of an object falling is:

$y(t)=y_{0}+v_{0y}t-\frac{1}{2}at^2$

But in this case I don't know how tall is the building neither of what else could I do or should I need to find what it is being asked.

I suspect that I should sum the vectors for the velocity of the car and that of the bucket falling and from then I could find the time. But I don't know which steps should I take to find such time. Can somebody help me here?.
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November 4th, 2019, 08:31 PM   #2
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$v_b = -10 t \, \hat{j} \, m/s$

$v_c = (15 \hat{i} - 15 \hat{j}) \, m/s$



$-10t \, \hat{j} - (15 \hat{i} - 15 \hat{j}) = -15\hat{i} \, m/s$

$-15\hat{i} - (10t - 15)\hat{j} = -15\hat{i} \, m/s$

$10t -15 = 0 \implies t = 1.5 \, sec$
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