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A toy rocket is launched vertically from ground level \((y = 0 \text{ m})\) at time \(t = 0.0 \text{ s}\). The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to \(64 \text{ m}\) and acquired a velocity of \(60 \text{ m/s}\). The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground.

What is the time interval during which the rocket engine provides upward acceleration?

Answer :

Final answer:

The time interval during which the rocket engine provides upward acceleration is approximately 9.88 seconds.

Explanation:

To find the time interval during which the rocket engine provides upward acceleration, we need to determine the time at which the engine burnout occurs. We can use the equation v = vo + at, where v is the final velocity, vo is the initial velocity, a is the acceleration, and t is the time. Given that the rocket has a final velocity of 60 m/s and an initial velocity of 0 m/s, and assuming constant acceleration, we can solve for the time t.

60 = 0 + at

60 = at

t = 60/a

Now we need to find the acceleration a. We know that the rocket has risen to a height of 64 m, so we can use the equation d = vot + (1/2)at^2, where d is the displacement, vo is the initial velocity, a is the acceleration, and t is the time. Substituting the known values, we get:

64 = 0*t + (1/2)at^2

64 = (1/2)at^2

128 = at^2

a = 128/t^2

Substituting the value of a into the equation for t, we get:

t = 60/(128/t^2)

t^3 = 60*128

t = ∛(60*128)

t ≈ 9.88 s

Therefore, the time interval during which the rocket engine provides upward acceleration is approximately 9.88 seconds.

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