Arescue plane wants to drop supplies to isolated mountain climbers on a
rocky ridge 235m below?
part (a)
if the plane is traveling horizontally with a speed of 61.1m/s, how far in
advance of the recipients (horizontal distance) must the goods be dropped?
part (b)
suppose, instead, that the plane releases the supplies a horizontal distance
of 425m 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?

Part (a): The goods must be dropped 422.812 meters advance of the recipients.

Part (b): The supplies must be given with ascendent vertical velocity of V=0.147 m/s to make they arrive precisely at the climbers position.

Explanation:

h= 235m

g= 9.8 m/s²

V= 61.1 m/s

Part a:

fall time of supplies:

t= 6.92 sec

d= V*t

d= 61.1 m/s * 6.92 s

d= 422.812 m

Part b:

d= 425m

d=V*t

t=d/V

t= 6.95 sec

Δt= 6.95 sec - 6.92 sec

Δt=0.03 sec

The supplies must be given ascendent vertical velocity to compensate that difference of time Δt. the half of the difference must be used to the ascendent part and the other half will be used to descendent part of the supplies.

To calculate the vertical velocity:

Vo= g * Δt/2

Vo= 9.8 m/s² * (0.03 sec/2)

Vo= 0.147 m/sec

amplitude

hope this