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DYNAMICS PROBLEMS
  1. In the case where the helicopter is hovering and has no forward velocity, the range is given by
    R = Ulauncht
    Similarly, balancing force in the vertical direction gives:
    F = may
    where ay = g. Also,
    (2) v⌠⌡vodv = t⌠⌡0aydt
    Thus,
    (3) v(t) = vo + gt
    Height from which the bomb is dropped is given by
    (4) ⌠⌡dh = ⌠⌡vdt = ⌠⌡(vo + gt)dt
    By integrating the above equation results in
    (5) H = vot + (1)/(2)gt2
    Substituting vo = 0, since the initial velocity of the bomb along the y-axis is zero,
    (6) H = (1)/(2)gt2
    Distance covered by the bomb in the horizontal direction (Range) is given by R = Ulauncht, where t is the time of flight. The value of t is obtained using the expression for distance covered in the y-direction as follows:
    H = (1)/(2)gt2
     ⇒ tflight = √((2H)/(g))
    Now range is obtained as:
    R = Ulaunchtflight = Ulaunch√((2H)/(g)) = 140m
    figure images/proj_inc.PNG
    Figure 4 Trajectory of the bomb which has been launched at 45 degree by the helicopter in forward flight Ucruise = 200 km/h with an initial forward velocity Ulaunch = 50 km/h.
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