A missile is shot straight up into the air. At the peak of its trajectory, it breaks up into three pieces of equal mass, all of which move horizontally away from the point of the explosion. one piece travels in a horizontal direction of 28degrees east of north with a speed of 30 m/s. The second piece travels in a horizontal direction of 12degrees south of west with a speed of 8 m/s. What is the velocity of the remaining piece? Give both speed and angle.

Use the fact that the sum of the three momentum vectors, of the separate pieces, is zero.

Show your work if you need further assistance.

25.6 m/s

To solve this problem, we can use vector addition to find the velocity of the remaining piece.

Let's break down the information given:

Piece 1:
Direction: 28 degrees east of north
Speed: 30 m/s

Piece 2:
Direction: 12 degrees south of west
Speed: 8 m/s

To find the velocity of the remaining piece, we need to find the combined effect of Piece 1 and Piece 2. To do that, we can break down their velocities into horizontal and vertical components.

Piece 1 horizontal component = 30 m/s * sin(28)
Piece 1 vertical component = 30 m/s * cos(28)

Piece 2 horizontal component = 8 m/s * cos(12)
Piece 2 vertical component = -8 m/s * sin(12) (note the negative sign denotes south direction)

Now, add the horizontal and vertical components separately to find the overall horizontal and vertical components for the remaining piece:

Horizontal component = (Piece 1 horizontal component) + (Piece 2 horizontal component)
Vertical component = (Piece 1 vertical component) + (Piece 2 vertical component)

Finally, we can use these horizontal and vertical components to find the magnitude of the velocity and its direction:

Speed = square root of (Horizontal component^2 + Vertical component^2)
Angle = arctan(Vertical component / Horizontal component)

Now let's calculate these values:

Piece 1 horizontal component = 30 m/s * sin(28) = 30 * 0.469 = 14.07 m/s
Piece 1 vertical component = 30 m/s * cos(28) = 30 * 0.884 = 26.52 m/s

Piece 2 horizontal component = 8 m/s * cos(12) = 8 * 0.978 = 7.82 m/s
Piece 2 vertical component = -8 m/s * sin(12) = -8 * 0.208 = -1.67 m/s

Horizontal component = 14.07 m/s + 7.82 m/s = 21.89 m/s
Vertical component = 26.52 m/s - 1.67 m/s = 24.85 m/s

Speed = square root of (21.89^2 + 24.85^2) = square root of (479.5721 + 617.0225) = square root of (1096.5946) = 33.10 m/s (approx)
Angle = arctan(24.85 m/s / 21.89 m/s) = arctan(1.135) = 48.27 degrees (approx)

Therefore, the velocity of the remaining piece is approximately 33.10 m/s at an angle of 48.27 degrees.

To find the velocity of the remaining piece, we need to consider the conservation of momentum. Momentum is a vector quantity and is conserved in an isolated system.

Let's denote the mass of each piece as m and the initial velocity of the missile as v.

Since the missile breaks up into three pieces of equal mass, each piece will have a mass of m/3.

Now, let's break down the initial velocity of the missile into its vertical and horizontal components.

The vertical component of the initial velocity is 0 because the missile is shot straight up into the air.

The horizontal component of the initial velocity remains the same for all three pieces. Let's denote it as Vx.

Now let's calculate the horizontal component of the initial velocity, Vx.

Vx = v * cos(θ)

Since the missile is shot straight up into the air, the angle between the initial velocity and the horizontal direction is 90 degrees.

Therefore, Vx = v * cos(90°) = v * 0 = 0 m/s

Now, let's find the vertical component of the initial velocity for each piece:

Piece 1: Vertical component = v * sin(θ1) = v * sin(28°)
Piece 2: Vertical component = v * sin(θ2) = v * sin(-12°)

Now, since momentum is conserved, the sum of the vertical components of the velocity for all three pieces should be zero:

0 = Piece 1: Vertical component + Piece 2: Vertical component + Piece 3: Vertical component

0 = v * sin(28°) + v * sin(-12°) + Piece 3: Vertical component

From this, we can solve for the vertical component of the velocity for the remaining piece (Piece 3):

Piece 3: Vertical component = -v * sin(28°) - v * sin(-12°)

Since we have the horizontal and vertical components of the remaining piece's velocity, we can find its speed and angle.

The speed (v3) can be calculated using the Pythagorean theorem:

v3 = √(Piece 3: Vertical component^2 + Piece 3: Horizontal component^2)

v3 = √((-v * sin(28°) - v * sin(-12°))^2 + (0 m/s)^2)

Finally, the angle (θ3) can be found using inverse trigonometric functions:

θ3 = tan^(-1)(Piece 3: Vertical component / Piece 3: Horizontal component)

θ3 = tan^(-1)((-v * sin(28°) - v * sin(-12°)) / 0)

Please note that the angle calculation may not be possible if the horizontal component is zero. In such cases, the remaining piece would have a vertical velocity only, and the angle of the velocity would be either 90 degrees (upward) or -90 degrees (downward).