# Answer to Question #3865 in Mechanics | Relativity for mia

Question #3865

A plane is flying in a straight line at an unchanging height and unchanging speed. The pilot must drop an object onto a target in front of the plane. At what angle must the pilot see the target at the moment of dropping the object? At that moment, what is the distance between the target and the plane’s projection onto the ground? Disregard air resistance.

Expert's answer

Let H be the height of the flight and V be the speed of the plane.

Let S be the distance between the target and the plane’'s projection onto the ground.

As the pilot drops the object, it covers the height H during the time t:

H=(g*t

So t = sqrt(2*H/g).

During the time t, the object covers the distance S along the x-axis:

S = V*t.

S = V*sqrt(2*H/g).

The pilot sees the target at the angle A with the horizont:

A = arctan(H/S) = aarctan (H/(V*sqrt(2*H/g))).

Let S be the distance between the target and the plane’'s projection onto the ground.

As the pilot drops the object, it covers the height H during the time t:

H=(g*t

^{2})/2.So t = sqrt(2*H/g).

During the time t, the object covers the distance S along the x-axis:

S = V*t.

S = V*sqrt(2*H/g).

The pilot sees the target at the angle A with the horizont:

A = arctan(H/S) = aarctan (H/(V*sqrt(2*H/g))).

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