Respuesta :
The motion of the package can be described as the motion of a projectile,
given that it has an horizontal velocity and it is acted on by gravity.
- a) [tex]\vec{v}[/tex] = 70·i
- b) The package will reach ground in approximately 12.77 seconds.
- c) The speed of the package as it lands is approximately 145.51 m/s.
- d) The path of the package based on a stationary frame of reference is parabolic
- e) The path of the package as seen from the plane is directly vertical downwards
Reasons:
Velocity of the aircraft = 70 m/s
Direction of flight of the aircraft = Eastward
Height from which the aircraft drops the package, h = 800 m
a) The initial velocity of the package, [tex]\vec{v}[/tex] = 70·i
b) The time it will take the package to reach the ground, t, is given by the formula;
[tex]\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}[/tex]
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
Therefore;
[tex]\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}[/tex]
Which gives;
[tex]\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}[/tex]
The time it will take the package to reach the ground, t ≈ 12.77 seconds
c) The vertical velocity just before the package reaches the ground, [tex]v_y[/tex], is given as follows;
[tex]v_y^2[/tex] = 2·g·h
Therefore;
[tex]v_y[/tex] = √(2·g·h)
Which gives;
[tex]v_y[/tex] = √(2 × 9.81 × 800) ≈ 125.28
[tex]v_y[/tex] ≈ 125.28 m/s
Which gives; [tex]\vec{v}[/tex] = 70·i - 125.28·j
Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51
The speed of the package as it lands, |v| ≈ 143.51 m/s
d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.
Therefore;
The path of the package is the path of a projectile, which is a parabolic shape.
e) As seen by someone on the aeroplane, the horizontal velocity will be
zero, therefore, the package will appear as accelerating directly vertical
downwards.
Learn more about projectile motion here:
https://brainly.com/question/1130127
