Question

A rescue 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?

Answer 1

Answer:

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:

$t=\sqrt{\frac{2.h}{g} }$

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

Answer 2

Part (a) The horizontal distance in advance the goods should be dropped is approximately 422.812 meters

Part (b) The velocity to be given to the supplies is approximately 0.394 m/s up

The reason the above values are correct is as follows:

The known parameters are;

The vertical distance of the climbers below the airplane, h = 235 m

The horizontal velocity of the airplane, v = 61.1 m/s

Acceleration due to gravity, g ≈ 9.81 m/s²

Part (a) Required:

The distance in advance of the recipients the goods must be dropped

Solution:

The time, t, it will take the goods to drop is given by the following formula for free fall;

$t =\mathbf{ \sqrt \dfrac{2 \times h }{g }}$

$t =\ \sqrt {\dfrac{2 \times 235 }{9.81 } } \mathbf{\approx 6.92}$

The time it will take the goods to drop, t ≈ 6.92 seconds

The distance in advance the goods should be dropped, d, is given as follows;

Distance , d = Velocity, v × Time, t

∴ d = Velocity of the airplane, v × The time it will take the goods to drop, t

Which gives;

d ≈ 61.1 m/s × 6.92 s = 422.812 meters

The horizontal distance in advance the goods should be dropped, d ≈ 422.812 meters

Part (b) Given:

The horizontal distance in advance the airplane releases the supplies, d = 425 meters

Required:

The vertical velocity (up or down) to be given the to supplies so they arrive at the climbers position

Solution:

The time the  supplies spend in the air, t, is given as follows;

t = d/v

Where;

v = The horizontal velocity of the airplane  = 61.1 m/s

∴ t = 425 m/(61.1 m/s) =  6.95581015 s ≈ 6.96 s

The vertical distance of travel, s, is given by the following kinematic equation of motion

s = u·t + (1/2)·g·t²

Where;

u = The vertical velocity downwards

s = The distance = 235 meters (below)

g = +9.81 m/s² (downward motion)

Plugging in the values gives;

235 = u·6.96 + (1/2)×9.81×6.96² = 6.96·u + (1/2)×9.81×6.96 = 237.606048

u·6.96 = 237.606048- 240.345 = -2.738952

u = -2.738952/6.96 ≈ -0.394

The velocity to be given to the supplies, u ≈ -0.394 m/s down ≈ 0.394 m/s up (wards)

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