# Physics

posted by LINA

Also got lost on this question:

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4 m/s), (a) how far in advance of the recipients (horizontal distance) must the goods be dropped. (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. (c) with what speed do the supplies land in the latter case?

1. Damon

The package has the same horizontal velocity component as the plane. (Note- if you drop a bomb at low altitude, turn your airplane immediately after release)
Fall of 235 m
235 = (1/2)g t*2 = 4.9 t^2
t = 6.93 seconds to fall
horizontal problem
d = u t
d = 69.4 * 6.93 = 481 meters

==========================
if at 425 meters then t = 425m/69.4 m/s
= 6.12 seconds instead of 6.93
we better give it a negative Vi so it gets down faster

-235 = Vi (6.12) - 4.9 (6.12^2)
-235 = 6.12 Vi -184
-51.2 = 6.12 Vi
Vi = - 8.37

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