A rescue plane wants to drop supplies to isolated mountain climbers on a rocky r

Question

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. Assume the plane is traveling horizontally with a speed of 215 km/h (59.7 m/s).

(a) How far in advance of the recipients (horizontal distance) must the goods be dropped (Figure 3-37a)-413.4meters

(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 (Figure 3-37b)? (Assume up is positive.)?

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

I am stuck on parts C and B.

Explanation / Answer

a) First let's figure out the time required to fall 235 m with 0 initial vertical speed. d = 1/2 at^2 t = sqrt(2d/a) = sqrt(2x235/9.81) = 6.9s In 6.9s, the plane will travel 6.9s x 59.72m/s = 412.08m So the package has to be dropped 412.08m before the target. b) If the package is dropped at a later time (when d = 425 m), then it will need an additional downward speed to reach the target in time. Now t = 425/59.72 = 7.12s and d = 1/2at^2 + vt 235 = 1/2 (9.81)(7.12)^2 + v(7.12) v = -1.88 m/s c) v = at+vi = 9.81*7.12 -1.88 = 67.89m/s

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