Abstract

    In problem 2.10, we are going to simulate the best solution to reach a specific target.

    I set the target point at (100,3). 100 means that the distance between the lauch point and the target point is 100km. And 3 means the height.

Backgroud

    Based on the last problem I have worked. I use the Adiabatic approximation to simulate this time.

Main Body

    In exercise5, I have discussed air drug force, and altitude problem. So a few modifications can

    be suitable for this new situation.

    I changed some details in the code so that I can set the landing height is 3 kilometers.

    And last time I found the best shooting angle is 50.

    To reach the distance of 100km, I change the volecity to approach it.

    Here is the result and the code_6.1

For more accurate result, I chose the smaller step between 1620 and 1650, here is what we get

velocity    distance       time        velocity    distance      time      velocity    distance       time

1620 98.5371512027 163.35    1621 98.6505895603 163.44    1622 98.7620894555 163.525

1623 98.8736661569 163.61    1624 98.9873399262 163.7      1625 99.0990715454 163.785

1626 99.2108799719 163.87    1627 99.3247891627 163.96    1628 99.4367525101 164.045

1629 99.5487926651 164.13    1630 99.6629372849 164.22    1631 99.7751323623 164.305

1632 99.8894343001 164.395   1633 100.0017843 164.48        1634 100.114211107 164.565

1635 100.228748481 164.655   1636 100.34133021 164.74      1637 100.453988742 164.825

1638 100.568761553 164.915   1639 100.681575005 165.0      1640 100.794465259165.085

1641 100.909473505 165.175   1642 101.022518673 165.26    1643 101.135640639 165.345

1644 101.250884316 165.435   1645 101.364161191 165.52    1646 101.477514859 165.605

1647 101.592993961 165.695   1648 101.70650253 165.78      1649 101.822138942 165.87

So ,the precise velocity is 1633m/s

To confirm the the firing angle, I modified the code foranother time to scan the best firing angle at

the velocity of 1633m/s.

And here is the result and code6.2

angle           distance              time

45        94.7750083648        148.685

46        96.0868163794        151.91

47        97.2768219318        155.11

48        98.3304016304        158.27

49        99.2417834456        161.395

50        100.0017843             164.48

51        100.602011483          167.52

52        101.038927851          170.52

53        101.301805647          173.465

54        101.390788905          176.365

55        101.298907246          179.21

56        101.024002272          182.0

Conclusion

After problem2.9 and problem2.10, to attack a target which is 3 km higher than the firing ponit and 1000 km from the cannon,we should fire at the velocity of 1633 m/s and the angle of 50 degree.

Reference

This work is completely finished by my own.