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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 $10 \sqrt 33$

Vedant Chourey
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Robert
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  • You're derivatives aren't quite right; their magnitudes are right, but their signs are wrong. It looks like the answer $10\sqrt{33}$ is incorrect. – rnrstopstraffic Mar 23 '15 at 16:36
  • Also, have you written this out using functions? @john has the right idea that since the unit vectors are orthogonal, we can treat them independently. Thus, the best approach would be to build your velocity function by integrating the acceleration function (using your initial velocity). The position function can be found by integrating the velocity function (using your initial position). If done this way, you've gained all of the information in a usable form. – rnrstopstraffic Mar 23 '15 at 16:39
  • Oh okay, yeah I was thinking it was probably the answer that was wrong. No, I should try that now that I think of it. Thanks. – Robert Mar 23 '15 at 16:43

1 Answers1

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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:

$$v = \sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = \sqrt{9300} ft/s.$$

John
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