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26 July, 22:45

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 39.2 m/s^2. The acceleration period lasts for time 10.0 ss until the fuel is exhausted. After that, the rocket is in free fall.

Required:

Find the maximum height ymax reached by the rocket.

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Answers (1)
  1. 26 July, 22:48
    0
    9800 m

    Explanation:

    During acceleration, given:

    v₀ = 0 m/s

    a = 39.2 m/s²

    t = 10.0 s

    Find: v and Δy

    v = at + v₀

    v = (39.2 m/s²) (10.0 s) + 0 m/s

    v = 392 m/s

    Δy = v₀ t + ½ at²

    Δy = (0 m/s) (10.0 s) + ½ (39.2 m/s²) (10.0 s) ²

    Δy = 1960 m

    During free fall, given:

    v₀ = 392 m/s

    v = 0 m/s

    a = - 9.8 m/s²

    Find: Δy

    v² = v₀² + 2aΔy

    (0 m/s) ² = (392 m/s) ² + 2 (-9.8 m/s²) Δy

    Δy = 7840 m

    Therefore, h = 1960 m + 7840 m = 9800 m.
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