Question

A helicopter is flying in a straight line over a level field at a constant speed of 12.0 m/s and at a constant altitude of 9.50m/s. A package is ejected horizontally from the helicopter with an initial velocity of 12.0m/s relative to the helicopter and in a direction opposite the helicopter’s otion.

(a) Find the initial speed of the package relative to the ground.
(b) What is the horizontal distance between the helicopter and the package at the instant the package strikes the ground?
(c) What angle does the velocity vector of the package make with the ground at the instant before impact, as seen from the ground?

Answer 1

The initial speed of the package relative to the ground is 12.0 m/s. The horizontal distance between the helicopter and the package at the instant the package strikes the ground is 12.0 m/s multiplied by the time of flight. The angle that the velocity vector of the package makes with the ground at the instant before impact can be found using the inverse tangent function.

Step-by-step explanation:

(a) To find the initial speed of the package relative to the ground, we need to consider the velocity of the helicopter and the initial velocity of the package relative to the helicopter. Since the package is ejected horizontally in the direction opposite to the helicopter's motion, the horizontal component of the package's velocity relative to the ground is equal to the horizontal component of the helicopter's velocity. Therefore, the initial speed of the package relative to the ground is 12.0 m/s.

(b) To find the horizontal distance between the helicopter and the package at the instant the package strikes the ground, we can use the formula:

distance = speed × time

Since the speed of the package relative to the ground is the same as the speed of the helicopter, and the time of flight of the package is the same as the time of flight of the helicopter, the horizontal distance traveled by the package is the same as the horizontal distance traveled by the helicopter. Therefore, the horizontal distance between the helicopter and the package at the instant the package strikes the ground is given by:

distance = speed × time = 12.0 m/s × time

(c) To find the angle that the velocity vector of the package makes with the ground at the instant before impact, we can use trigonometry. The vertical component of the velocity vector remains constant at 9.50 m/s, and the horizontal component of the velocity vector is equal to the horizontal component of the helicopter's velocity. Using the inverse tangent function, we can find the angle:

angle = arctan(vertical velocity / horizontal velocity) = arctan(9.50 m/s / 12.0 m/s)