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The question describes a projectile launched from ground at an angle of $40$ degrees with an initial velocity of $96$ ft/s. I understand the concept of resolving this $96$ into x and y components. So, I can easily answer the questions of Max height (59.5 ft), how long will it take for the thing to hit the ground ($3.86 s$) and the range or max horizontal distance ($283.6$ ft). However, it then asks for the impact velocity. I now see how the $V_(xf)=V_(xi)$ since the $a_x=0$ and how $V_(yf)=V_(yi)$, however I don't see how this angle is $140$ degrees. I do see how the $V_f$ is negative, but I don't see how we can just arbitrarily assign $140$ degrees to make the final velocity negative. How do I mathematically show the angle is in Q2?

How do I get the impact angle?

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  • $\begingroup$ For an object launched vertically upwards, what is the relation between, at the same horizontal height, of the object while going upwards, and coming back down? $\endgroup$ – AnotherJohnDoe Sep 30 '17 at 21:03
  • $\begingroup$ It would seem to me that the x direction is to the right and the vertical direction at impact is downward, so that would put the angle in Q4 and the angle would be 320 not 140. I don't see how this angle becomes 140. $\endgroup$ – user163862 Sep 30 '17 at 21:08
  • $\begingroup$ I'm confused in your question about horizontal height. Did you mean vertical height? $\endgroup$ – user163862 Sep 30 '17 at 21:09
  • $\begingroup$ Oops, I meant at a particular vertical height $\endgroup$ – AnotherJohnDoe Sep 30 '17 at 21:10
  • $\begingroup$ the theta would be the same $\endgroup$ – user163862 Sep 30 '17 at 21:29
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The impact speed is just the launch speed (by energy conservation), and the angle is $-40^\circ$. Because the $x$ component velocity is the same, and the $y$ component reversed. However, if you define the impact velocity as the angle between the trajectory with the +ve $x$ axis at the moment when it hits the ground, then $140^\circ$ is also ok (if it is from a physics textbook :) ). But I would say $-40^\circ$ (or $320^\circ$) is more correct mathematically.

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  • $\begingroup$ YESSS....I would think the angle would be $320$ or $-40$, NOT $140$ $\endgroup$ – user163862 Sep 30 '17 at 21:35
  • $\begingroup$ It would be a Q4 angle. $\endgroup$ – user163862 Sep 30 '17 at 21:35
  • $\begingroup$ No, it's from a calculus class. $\endgroup$ – user163862 Sep 30 '17 at 21:37
  • $\begingroup$ Mathematically, it should be $-40^\circ$ or $320^\circ$. But I think $140^\circ$ is an acceptable answer in physics. $\endgroup$ – velut luna Sep 30 '17 at 21:37
  • $\begingroup$ Why does 140 work in physics? $\endgroup$ – user163862 Sep 30 '17 at 21:37

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