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A toy rocket is launched vertically from ground level at time [tex]t = 0.00 \, \text{s}[/tex]. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 64 m and acquired an upward velocity. The rocket continues to rise with insignificant air resistance in unpowered flight, reaches maximum height, and falls back to the ground.

The time interval during which the rocket engine provided the upward acceleration is closest to:

[Provide options or additional information as necessary for context]

Answer :

Final answer:

The question relates to the time of upward acceleration of a rocket. It involves applying equations of motion to find the acceleration as per the given conditions and then determining the time taken during the burn phase of the rocket.

Explanation:

This question involves the concept of a rocket's upward acceleration. From the given details, the rocket has an initial vertical velocity of 0 (since it is launched from the ground). It rises to a height of 64 meters and acquires an upward velocity when running out of fuel (end of burn phase). After this, it follows a basic projectile motion.

To find the time during which the rocket engine provided the upward acceleration, we can apply the equation of motion: v² = v₀² + 2a(y - y₀), where v is the final velocity (acquired at end of burn phase), v₀ is the initial velocity (which is 0 in this case), a is acceleration, y is the final height (64m), and y₀ is the initial height (0 in this case). Solving the equation for 'a' the acceleration, we can find the time taken using the equation v = v₀ + at.

This exercise represents a real-world application of equations of motion and the concept of acceleration in the context of rocket science.

Learn more about Rocket Acceleration here:

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