A ball is thrown eastward into the air from the origin (in the direction of the positive x − axis ). The initial velocity is $50i + 80k$. The spin of the ball results in a southward acceleration of $4 ft /s^2$ , so the acceleration vector is $\overrightarrow{a} = -4j - 32k$ . Where does the ball land and with what speed?
I found the answer to first part of the question, which is $<250, -50, 0>$ (I am sure this is correct), but I don't understand how to arrive at the second part.
So far, I have calculated $t = 5$ seconds
$v_xf = 50$
$v_yf = 20$
$v_zf = 80$
and arrive at the speed $\sqrt 9300$, but the answer is $10 \sqrt 33$
Treat each coordinate independently (since the coordinates are orthogonal and there's no coupling).
After $5$ seconds (when the ball lands), the $x$ component of the velocity will be the same: $50 ft/s$. The $z$ component will be the negative of its starting velocity: $-80 ft/s$. The $y$ component will be its acceleration over that time, times the time: $-4 \cdot 5 = -20 ft/s$.
Then, taking the square root of the sum of squares:
$$v = \sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = \sqrt{9300} ft/s.$$