Spymaster Paul, flying a constant 215km/h horizontally in a low-flying helicopter, want to drop secret documents into his contact’s open car which is traveling 155km/h on a level highway 78.0m below. At what angle (to the horizontal) should the car be in his sights when the package is released?

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aristocles aristocles · Answer 1 · 2019-09-13T02:37:59+02:00

Answer:

[tex]\theta = 49.56 degree[/tex]

Explanation:

Relative horizontal velocity of plane with respect to the car is given as

[tex]v_r = v_p - v_c[/tex]

so we have

[tex]v_p = 215 km/h[/tex]

[tex]v_c = 155 km/h[/tex]

now we have

[tex]v_r = 215 - 155 [/tex]

[tex]v_r = 60 km/h[/tex]

[tex]v_r = 16.67 m/s[/tex]

Now time taken by the object to drop the vertical height is given as

[tex]y = \frac{1}{2}gt^2[/tex]

[tex]78 = \frac{1}{2}(9.81)t^2[/tex]

[tex]t = 3.98 s[/tex]

so the distance of the car must be

[tex]d = v_r\times t[/tex]

[tex]d = 16.67 \times 3.98[/tex]

[tex]d = 66.47 m[/tex]

Angle of the car with horizontal is given as

[tex]tan\theta = \frac{y}{x}[/tex]

[tex]tan\theta = \frac{78}{66.47}[/tex]

[tex]\theta = tan^{-1}(1.17)[/tex]

[tex]\theta = 49.56 degree[/tex]