Answer:
a. The speed of the helicopter at the moment the package is dropped is approximately 15.519 m/s
b. The position of the package 2s after it is dropped is (31.038 m, -9.8 m)
c. The time of flight of the food package is approximately 5.105 seconds
d. The horizontal distance covered is approximately 79.22 m
Explanation:
The given parameters are;
The height at which the helicopter is flying = 125 m
The magnitude of the velocity of the package after 2 seconds, v = 25 m/s
a. The vertical velocity of the package, [tex]v_y[/tex], is given as follows;
[tex]v_y[/tex] = [tex]u_y[/tex] + g·t
Where;
[tex]u_y[/tex] = 0 m/s for the dropped food package
t = The duration of the package in the air = 2 s
g = The acceleration due to gravity = 9.8 m/s²
Therefore;
[tex]v_y[/tex] = g·t = 9.8 m/s² × 2 s = 19.6 m/s
We have, [tex]v = \sqrt{v_x^2 + v_y^2}[/tex],
Where;
vₓ = The horizontal velocity of the food package = The horizontal speed of the helicopter
therefore, [tex]v_x = \sqrt{v^2 - v_y^2}[/tex], which gives;
[tex]v_x = \sqrt{v^2 - v_y^2} = \sqrt{25^2 - 19.6^2} \approx 15.519 \ m/s[/tex]
The speed of the helicopter at the moment the package is dropped ≈ 15.519 m/s
b. The vertical if the package after 2 s is given as follows;
The vertical position of the package after 2 s is given by the kinematic equation, h = 1/2·g·t² = 1/2 × 9.8 m/s² × 2 s = 9.8 m
The food package is located 9.8 m vertically below the point where it is dropped
The horizontal position, x, of the package after 2 s is given by the following kinematic equation, x = vₓ × t
Therefore, x = 15.519 m/s × 2 s ≈ 31.038 m
The coordinates of the location of the package 2 seconds after it is dropped is (31.038 m, 9.8 m)
c. The time of flight, [tex]t_{tot}[/tex], of the food package which is the time it takes the package to reach the ground from 125 m, is given as follows;
[tex]t_{tot} = \sqrt{\dfrac{2 \cdot h}{g} } = \sqrt{\dfrac{2 \times 125 \ m}{9.8 \ m/s^2} } \approx 5.051 \ seconds[/tex]
The time of flight of the food package, [tex]t_{125}[/tex] ≈ 5.105 seconds
d. The horizontal distance covered, [tex]x_{tot}[/tex] = vₓ × [tex]t_{tot}[/tex] ≈ 15.519 m/s × 5.105 s = 79.224495 m ≈ 79.22 m.
