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An emergency relief plane is dropping a care package from a plane to a group of medical personnel working for a relief agency in an African village. The package is designed to land in a small lake, inflate an attached raft upon impact, and finally resurface with the raft side down. The plane will be moving horizontally with a ground speed of 59.1 m/s. The package will be dropped a horizontal distance of 521 m from the intended target location. At what altitude above the pond must the plane be flying in order to successfully accomplish this feat?

Sagot :

Numenius

Answer:

The altitude of the plane is 379.5 m.

Explanation:

Initial horizontal velocity, u = 59.1 m/s

Horizontal distance, d = 521 m

let the time taken by the packet to cover the distance is t.

Horizontal distance = horizontal velocity x time

521 = 59.1 x t

t = 8.8 s

let the vertical height is h .

Use second equation of motion in vertical direction.

[tex]h = u t + 0.5 gt^2\\\\h = 0 + 4.9 \times 8.8\times8.8\\\\h= 379.5 m[/tex]

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