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salantis [7]
3 years ago
9

Help meeeee

Physics
2 answers:
dezoksy [38]3 years ago
7 0

You just have to simpy add the vectors together.  Draw a picture to help you.  Your verticle run is a vector that has a magntiude of 22m along the x axis with an angle of 0, and your climb has is a vector with a magnitude of 4.8 along the y axis and an angle of 90 degreed with respect to the x axis therefore add all of the x components and y components.  you will find that the resultant vector has a xcomponent of 22m and a y component of 4.8m so draw a graph the point will be (22,4.8) and then you can simply measure the angle.

Umnica [9.8K]3 years ago
7 0

It's going to be the angle whose tangent is (4.8 / 22).  That's about 0.2182, and you can easily use your calculator to find the angle that has that tangent.

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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, v_y, of the package just before landing is given by the relation;

v_y² = u² + 2·g·h

u = 0 m/s

∴ v_y² = 0 + 2×9.81×1500 = 29430 m²/s²

v_y = √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

∴ θ = tan⁻¹(4.4) = 77.19°.

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