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Spymaster Tim, flying a constant vh = 193.6 km/hr horizontally in a low-flying helicopter, wants to drop secret documents into his contact's open car which is traveling va = 141.7 km/hr on a level highway h = 71.3 m below. At what angle (with the horizontal) should the car be in his sights when the packet is released?

Respuesta :

Answer:

θ = 51.21°

Explanation:

Vh= 193.6 km/h

Va= 141.7 km/hr

Relative velocity ,Vr= 193.6 - 141.7 = 51.9 km/hr

Time taken to cover 71.3 m

[tex]h=\dfrac{1}{2}gt^2[/tex]

[tex]71.3=\dfrac{1}{2}\times 9.81\times t^2[/tex]

t= 3.81 s

t= 0.00105 hr

So the horizontal distance x

x= Vr .t

x= 51.9 x 0.00105  km

x=0.0549 m

x= 54.96 m

Lets take angle is θ

tan θ = h/x

tan θ = 71.3/54.96

θ = 51.21°