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A toy rocket is launched vertically from ground level at time \( t = 0.00 \, \text{s} \). The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to \( 81 \, \text{m} \) and acquired an upward velocity of \( 40 \, \text{m/s} \). The rocket continues to rise with insignificant air resistance in unpowered flight, reaches maximum height, and falls back to the ground.

What is the upward acceleration of the rocket during the burn phase?

Answer :

The required upward acceleration would be approximately 9.88 m/s^2

How to find the upward acceleration

To find the upward acceleration during the burn phase, you can use the kinematic equation:

v^2 = u^2 + 2 * a * d

Here, v is the final velocity of 40 m/s, u is the initial velocity of 0 m/s (since it starts from rest), a is the acceleration, and d is the distance of 81 m.

Solve for a:

1600 = 0 + 2 * a * 81

1600 = 162 * a

a = 1600 / 162

a = 9.87654321 m/s^2

Rounded, the acceleration would be approximately 9.88 m/s^2.

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