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Time of range?

  1. Feb 20, 2005 #1
    During World War 1, the Germans had a gun called Big Bertha that was used to shell Paris. the shell had an initial speed of 1.22 km.s at an initial inclination of 85.3 degrees to the horizontal. The acceleration of gravity is 9.8 m/s^2. How far did the shell hit? Answer in units of km.
    i got that answer to be .0248055472 km and that was right then i tried
    visin(theta)/g=t, and i got 124.0711849 as my time, but that wasnt right? then what equation do i use? thank you
  2. jcsd
  3. Feb 20, 2005 #2
    You must use the horizontal component of the initial velocity, because you found the horizontal distance traveled. Use the angle value and trigonometry to find it.
  4. Feb 20, 2005 #3
    vxi=1.22cos(85.3)=.099649805, so then the distance is 24.8055472 m, and i still dont see how you use trig to get t?
  5. Feb 20, 2005 #4
    Watch your units and make sure you use either meters or kilometers consistently. I meant use trig to find the horizontal component of velocity, but I now see you've already done that. Now remember that

    [tex]v=\frac{\Delta d}{\Delta t}[/tex]

    and solve for [itex]t[/itex]. (Hint: this is a gun, and the distance is short; it's going to be quick.)
  6. Feb 20, 2005 #5
    d=.0248055472 which is right, is vxi = 1.22(vi)cos(85.3)? cause when i did .0248055472/.0999649805=.2481423702 wasnt right and i only have one more try
  7. Feb 20, 2005 #6
    both the 1.22 and the distance are in km
  8. Feb 20, 2005 #7
    Highly weird. That's the answer I get. The only thing I could think of is that this program or whatever you're using requires rules regarding significant digits to be respected, but your previous answer was accepted and it was unrounded, so I'm not sure. Perhaps someone else will see what we're missing.
  9. Feb 20, 2005 #8
    this is the answer it has recorded for 9) 0.0248055, i seriously think this program is off sometimes even a little bit cause ive gotten a lot of close answers but none that are exactly what it was on the computer
  10. Feb 20, 2005 #9
    but in your mind my answer is correct right? the .248141898 s? casue if it is im going to go into his office hours and show him that i had it right
  11. Feb 20, 2005 #10
    Yes, that is the answer I get. Explain your method to your prof, perhaps he knows something we don't, but this is how I would do the problem, and the answer seems reasonable.
  12. Feb 20, 2005 #11
    ill see what his answer is at midnight then get back to you so you can see how he did it
  13. Feb 20, 2005 #12
    someone suggested 0=1.22sin(85.3)-(9.8)(t), and i got .1240711849 which was my first answer
  14. Feb 20, 2005 #13
    correct answer was 248.142 how is that s, cause i had that answer but it was .248142, why did they turn the km in m, shouldnt it not have made a differnece?
  15. Feb 20, 2005 #14
    Hm. The suggestion in post #12 is incorrect because the horizontal component of the velocity of the projectile is constant and unaffected by gravitational acceleration.

    Alright, the answer listed was the same but with the decimal moved over, so don't worry too much. You are right, it should not make a difference if you use kilometers or meters as long as you are consistent throughout the problem. I'm not sure how they got that answer. I wonder what your prof will say about the method (hey, I'm learning too!).
  16. Feb 20, 2005 #15
    we all are but that is just what is making me is i got .248142 and the real answer according to the computer was 248.142
  17. Feb 20, 2005 #16
    no i messed up my answer for 9 was 24.8055 , cause it was in km not m and then divide by the vxi which was .0999649 and got the answer i should have gotten
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