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Rocket is fired from ground at a building 20,000 ft away. The initial velocity is $1,000 ft/s$ at an angle of $20^\circ$. How long will it take for the projectile to hit the building? I resolved the initial velocity triangle into the appropriate right triangle I use the equation $\Delta y=v_(iy)t + 1/2 at^2$. I get $t=21.375s$. Book gets $21.3$ so I assume that my answer is correct.

Then the question asks what will be the impact velocity. This is a math class. But in physics I was taught that not counting air friction, there would be a conservation of energy and the final velocity would be the initial velocity. So, I would think the answer would be $1000 f/s.$ The calculus book answer is $999.17$ at $160.13^\circ$.

Totally at a loss here as to how they get these numbers. I'm grasping at straws here, but I notice that the x direction of travel in 21.375 seconds is MORE than the 20,000 feet of the building. So, is the fact that the rocket hits the building before the 21.375 seconds factor into this? I've never encountered a problem like this.

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  • $\begingroup$ OK, so I recalculate the time to be $20000/v_x=21.3s$ which I now see is not the same as my answer of 21.375. It hits the building $.075$ sooner than if it had hit the earth. $\endgroup$ – user163862 Sep 30 '17 at 22:48
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The time of flight can be obtained by $$T=\frac{20000}{1000\cos 20^\circ}=21.284\text{ s}$$

The vertical velocity during impact is $$v_y(T)=v_y(0)-gT$$ $$=1000\sin 20^\circ -g\frac{20000}{1000\cos 20^\circ}$$

Taking $g=32\text{ ft/s}^2$, you have $$v_y(T)=-339.05363\text{ ft/s}$$

Of course, $v_x(T)=1000\cos 20^\circ=939.69262$

So

$$v(T)=\sqrt{939.69262^2+339.05363^2}=998.99\text{ ft/s}$$

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  • $\begingroup$ yes, I got the same as you in my comment above. but I still don't know how to get final velocity. I put it into the $v_f^2$ equation and it doesn't work $\endgroup$ – user163862 Sep 30 '17 at 22:50
  • $\begingroup$ I used $v_f^2=v_(ix)^2 +2ad$ and for d I used 20000 and $ a=32$ $\endgroup$ – user163862 Sep 30 '17 at 22:52
  • $\begingroup$ I get $1005$ which makes no sense that it would be MORE than the original. $\endgroup$ – user163862 Sep 30 '17 at 22:53
  • $\begingroup$ how do they get 999.17 at 160.13 deg? $\endgroup$ – user163862 Sep 30 '17 at 22:56
  • $\begingroup$ I don't know what value of $g$ you usually take. We usually take $9.8$ or $9.81$. $\endgroup$ – velut luna Sep 30 '17 at 22:59

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