Ridge Distance

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. The plane is?

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. The plane is traveling horizontally with a speed of 240 km/h (66.7 m/s).

Figure 3-41.
(a) How far in advance of the recipients (horizontal distance) must the goods be dropped (Fig. 3 41a)?
m
(b) Suppose, instead, that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers' position (Fig. 3 41b)?
Magnitude
m/s
Direction
up
left
down
right

(c) With what speed do the supplies land in the latter case?
m/s

s = 0.5*a*t^2 = 235m = 0.5*9.8m/s^2*t^2
t = sqrt(235/4.9) s = 6.9s
x = Vh*t = 66.7 m/s * 6.9s = 462m

b) At this range, the supplies will overshoot the target unless they are given an extra "push" down.
Now we have less time for freefall:
t = 6.9s - (462m-425m)/66.7m/s = 6.9s - 0.555s = 6.37s
s = Vo*t + 0.5*a*t^2 = 235m = Vo*6.37s + 0.5*9.8m/s^2*(6.37s)^2 = Vo*6.37s + 199 m
Vo = (235m-199m)/6.37s = 5.67 m/s