Answer:
(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane
(b) The horizontal component = 39.0 m/s
The vertical component = 171.55 m/s
(c) The angle of impact is 77.19°
Explanation:
The parameters given are;
Velocity of the plane, vₓ = 39.0 m/s
Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m
(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;
h = u·t + 1/2×g×t²
Where:
g = Acceleration due to gravity = 9.81 m/s²
u = Initial vertical velocity = 0 m/s
Hence;
1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²
∴ t = √(1500/4.905) = 17.49 s
The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m
The package lands 682 meters horizontally ahead from the point the package was dropped from the plane
(b) The vertical velocity,
, of the package just before landing is given by the relation;
² = u² + 2·g·h
u = 0 m/s
∴
² = 0 + 2×9.81×1500 = 29430 m²/s²
= √29430 = 171.55 m/s
Hence the horizontal component = 39.0 m/s
The vertical component = 171.55 m/s
(c) The angle of impact, θ, is given as follows;
![tan \theta = \dfrac{v_y}{v_x} = \dfrac{171.55}{39.0 } = 4.4](https://tex.z-dn.net/?f=tan%20%5Ctheta%20%3D%20%5Cdfrac%7Bv_y%7D%7Bv_x%7D%20%20%3D%20%5Cdfrac%7B171.55%7D%7B39.0%20%7D%20%3D%204.4)
∴ θ = tan⁻¹(4.4) = 77.19°.