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?.
If the car is moving with the velocity $15\hat i-15\hat j$ with respect to the ground, in the reference frame of the car, the bucket moves with velocity $-15\hat i+15\hat j$. There is no force in the $\hat i$ direction, just in the $\hat j$ direction. You can write $$v_i(t)=v_i(0)\\v_j(t)=v_j(0)-gt$$You have $v_j(t)=0$, $v_j(0)=15$ so $$0=15-10t$$or $t=1.5$ seconds.