A helicopter is ascending vertically with a speed of 5.13 m/s . At a height of 112 m above the Earth, a package is dropped from
1 answer:
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Answer: 3 m/s
Explanation:
We can solve the problem by using the law of conservation of momentum: during the collision between the two balls, the total momentum of the system before the collision and after the collision must be conserved:
The total momentum before the collision is given only by the cue ball, since the solid ball is initially at rest, therefore
So, the final total momentum will also be
And the total momentum after the collision is given only by the solid ball, since the cue ball is now at rest, therefore:
from which we find the velocity of the solid ball
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;
∴ θ = tan⁻¹(4.4) = 77.19°.