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

Question

A rescue plane wants to drop supplies to isolated mountain climbers or a rocky ridge 200m below. If the plane is traveling horizontally with a speed of 250 km/hr (69.0 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 400m in advance of the mountain climbers. What vertical velocity should they give the supplies (up or down) so that they arrive precisely at the climber's position? (c) With what speed do the supplies land in the latter case?

Explanation / Answer

(A) In vertical,

displacement of goods to reach, y = - 200 m
plane is travelling horizontally hence initial vertical velocity is zero.

v0y = 0

Applying y = v0y + a t^2 /2

- 200 = 0 - 9.8 t^2 /2

t = 6.39 sec

In horizontal d= v0x t

d = 69 x 6.39 = 440.8 m .........Ans

(B) time taken to travel 400 m horizontally,

t = 400 / 69 = 5.797 sec

IN vertical, - 200 = v0y (5.797) - 9.81 ( 5.797^2 ) /2

v0y = - 6.07 m/s

hence 6.07 m/s downward .........Ans

(C) vfx = v0x = 69 m/s

vfy = - 6.07 - (9.8 x 5.797) = - 62.94 m/s

Speed = sqrt( vfx^2 + vfy^2)

= 93.4 m/s

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