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2 November, 13:50

A hockey puck slides off the edge of a platform with an initial velocity of 20 m/s horizontally. The height of the platform above the ground is 2.0 m. What is the magnitude of the velocity of the puck just before it touches the ground

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  1. 2 November, 16:43
    0
    20.96 m/s

    Explanation:

    Using the equations of motion

    y = uᵧt + gt²/2

    Since the puck slides off horizontally,

    uᵧ = vertical component of the initial velocity of the puck = 0 m/s

    y = vertical height of the platform = 2 m

    g = 9.8 m/s²

    t = time of flight of the puck = ?

    2 = (0) (t) + 9.8 t²/2

    4.9t² = 2

    t = 0.639 s

    For the horizontal component of the motion

    x = uₓt + gt²/2

    x = horizontal distance covered by the puck

    uₓ = horizontal component of the initial velocity = 20 m/s

    g = 0 m/s² as there's no acceleration component in the x-direction

    t = 0.639 s

    x = (20 * 0.639) + (0 * 0.639²/2) = 12.78 m

    For the final velocity, we'll calculate the horizontal and vertical components

    vₓ² = uₓ² + 2gx

    g = 0 m/s²

    vₓ = uₓ = 20 m/s

    Vertical component

    vᵧ² = uᵧ² + 2gy

    vᵧ² = 0 + 2*9.8*2

    vᵧ = 6.26 m/s

    vₓ = 20 m/s, vᵧ = 6.26 m/s

    Magnitude of the velocity = √ (20² + 6.26²) = 20.96 m/s
  2. 2 November, 17:18
    0
    V = 20.96 m/s

    The magnitude of the velocity of the puck just before it touches the ground is 20.96 m/s

    Explanation:

    For the vertical component of velocity;

    Using the equation of motion;

    vᵥ^2 = u ^2 + 2as

    Where;

    vᵥ = final vertical speed

    u = initial vertical speed = 0

    (The puck only have horizontal speed.)

    a = g = acceleration due to gravity = 9.8m/s^2

    s = vertical distance covered = 2.0m

    initial horizontal speed = 20 m/s

    The equation becomes;

    vᵥ^2 = u ^2 + 2gs

    Substituting the given values, we have;

    vᵥ^2 = 0 + 2 (9.8*2)

    vᵥ^2 = 39.2

    vᵥ = √39.2

    vᵥ = 6.26 m/s

    For the horizontal component of velocity;

    Given that effect of the air resistance is negligible, then the final horizontal speed equals the initial horizontal speed since there is no acceleration.

    vₕ = uₕ = 20 m/s

    The resultant magnitude of velocity can be derived by;

    V = √ (vₕ² + vᵥ²)

    V = √ (20² + 6.26²)

    V = 20.96 m/s

    The magnitude of the velocity of the puck just before it touches the ground is 20.96 m/s
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