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 205 km/h (56.9m/s).

(a) How far in advance of the recipients (horizontal distance) must the goods be dropped (Figure a)?
____ 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 (Figure b)? (Assume up is positive.)
____ m/s

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

Explanation / Answer

vox = 205 = 56.9 m/s

y = -h = -235 m

ay = -g = -9.8 m/s^2

(a)

let T be time taken for supplies to reach recipients

along horizantal

X = vox*T ========> T = X/vox

along vertical

voy = 0

y = voy*T + 0.5*ay*T^2

-235 = 0 -(0.5*9.8*x^2/56.9^2)

X = 394 m

+++++++++++++++++++

(b)

along horizantal

T = X/vox = 425/56.9 = 7.5 s

along vertical

initial velocity = voy

acceleration ay = -g

y = -235 m

y = voy*T + 0.5*ay*T^2

-235 = voy*7.5 -(0.5*9.8*7.5^2)

voy = +5.42 m/s <<---------answer

(c)

vy = voy + ay*T

vy = 5.42-(9.8*7.5) = 68.08 m/s

vx = vox = 56.9

v = sqrt(vx^2+vy^2) = sqrt(56.9^2+68.08^2) = 88.73 m/s <<----answer

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