(b) A ball is thrown upward from the top of a 30 m tower with initial velocity 90 m/s at an angle O = 20° Find the time to reach t

Question

(b) A ball is thrown upward from the top of a 30 m tower with initial velocity 90 m/s at an angle
O = 20°
Find the time to reach the ground
(Smarks)
Find the magnitude and direction of the velocity at the moment of impact (5 marks)?​

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Latifah 3 years 2021-08-06T16:00:54+00:00 1 Answers 18 views 0

Answers ( )

  1. thienthi
    0
    2021-08-06T16:02:28+00:00
    Reply

    Answer:

    7.14 s

    93.2 m/s, 24.9° below the horizontal

    Explanation:

    Given in the y direction:

    Δy = -30 m

    v₀ = 90 m/s sin 20° ≈ 30.8 m/s

    a = -9.8 m/s²

    Find: t

    Δy = v₀ t + ½ at²

    -30 m = (30.8 m/s) t + ½ (-9.8 m/s²) t²

    4.9t² − 30.8t − 30 = 0

    t = [ 30.8 ± √((-30.8)² − 4(4.9)(-30)) ] / 2(4.9)

    t = 7.14 s

    Find: vᵧ

    v² = v₀² + 2aΔy

    vᵧ² = (30.8 m/s)² + 2 (-9.8 m/s²) (-30 m)

    vᵧ = -39.2 m/s

    The magnitude of the velocity is:

    v² = vₓ² + vᵧ²

    v² = (90 m/s cos 20°)² + (-39.2 m/s)²

    v = 93.2 m/s

    The direction of the velocity is:

    tan θ = vᵧ / vₓ

    tan θ = (-39.2 m/s) / (90 m/s cos 20°)

    θ = -24.9°

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