Please help.?

A kid kicks a ball in such a way that the successive displacements of the ball are 6.00 m to the north, 4.50 m northest, 2.10 m at 30.0 degree west of south.

Find out the resultant displacement covered by the ball from starting point till ball becomes stationary again ?

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  • 2 months ago
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    1st Kick: 6 m (N)

    2nd Kick: 4.50 m (NE) => 4.50 x cos45* E & 4.50 x sin45* N = 3.18 m (N) & 3.18 m (E)

    3rd Kick: 2.10 m (30* West of South) = 2.10 x cos30* (S) & 2.10 x sin30* (W) = 1.82 m (S) & 1.05 m (W)

    Thus Final displacement from starting point = [6m (N) + 3.18m (N) + 1.82m (S)] & [3.18m (E) + 1.05m (W)]

    =>Thus Final displacement from starting point = [7.36m (N)] & [2.13m (E)]

    =>Thus Final displacement from starting point = √[(7.36)^2 + (2.13)^2] = 7.66m

    & By tanθ = 2.13/7.36 = 0.29 = tan16.14*

    =>θ = 16.14* from North towards East

  • 2 months ago

    Displacement along north-south direction (north is positive)

    = (6.00 + 4.50 cos45° - 2.10 cos30°) m

    = 7.36 m

    Displacement along east-west direction (east is positive)

    = (4.5 sin45° - 2.1 sin30°) m

    = 2.13 m

    Magnitude of resultant displacement

    = √(7.36² + 2.13²) m

    = 7.66 m

    tanθ = 2.13/7.36

    θ = tan⁻¹(2.13/7.36)

    θ = 16.1°

    Direction of resultant displacement: N 16.1° E

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