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A missile is moving 1810 m/s at a 20.0° angle. It needs to hit a target 19,500 m away in a 32.0° direction in 9.20 s. What is the magnitude of the acceleration that the engine must produce?

User Torstenvl
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1 Answer

5 votes

Answer:

Aprox: 465.4 ms/2

Explanation:

To find the magnitude of the acceleration the engine must produce, we'll break down the missile's motion into its horizontal and vertical components.

Given:

Initial velocity (V₀) = 1810 m/s

Launch angle (θ) = 20.0°

Target distance (d) = 19,500 m

Target direction (φ) = 32.0°

Time of flight (t) = 9.20 s

First, let's find the horizontal and vertical components of the missile's initial velocity (V₀x and V₀y).

V₀x = V₀ * cos(θ)

V₀y = V₀ * sin(θ)

V₀x = 1810 m/s * cos(20.0°) ≈ 1712.4 m/s

V₀y = 1810 m/s * sin(20.0°) ≈ 618.8 m/s

Now, let's find the horizontal and vertical displacements (dx and dy) of the missile using the time of flight (t) and the target distance (d).

dx = d * cos(φ)

dy = d * sin(φ)

dx = 19,500 m * cos(32.0°) ≈ 16,402.8 m

dy = 19,500 m * sin(32.0°) ≈ 10,236.8 m

Next, let's calculate the acceleration needed in the x-direction (ax) and y-direction (ay) to cover the horizontal and vertical displacements in the given time.

ax = (2 * dx) / t²

ay = (2 * dy) / t²

ax = (2 * 16,402.8 m) / (9.20 s)² ≈ 391.7 m/s²

ay = (2 * 10,236.8 m) / (9.20 s)² ≈ 257.7 m/s²

Finally, to find the magnitude of the acceleration, we can use the Pythagorean theorem:

acceleration magnitude (a) = √(ax² + ay²)

a = √(391.7 m/s²)² + (257.7 m/s²)² ≈ 465.4 m/s²

Therefore, the magnitude of the acceleration that the engine must produce is approximately 465.4 m/s².

User Brian Schmitt
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