In a football game a kicker attempts a field goal. The ball remains in contact with the kicker’s foot for 0.0580 s, during which time it exp

Question

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker’s foot for 0.0580 s, during which time it experiences an acceleration of 376 m/s2. The ball is launched at an angle of 59.9° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.

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Khải Quang 3 years 2021-08-29T05:47:24+00:00 1 Answers 4 views 0

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  1. thienan
    0
    2021-08-29T05:48:47+00:00
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    Answer:

    V₀ₓ = 10.94 m/s

    V₀y = 18.87 m/s

    Explanation:

    To find the launch velocity, we use 1st equation of motion.

    Vf = Vi + at

    where,

    Vf = Final Velocity of Ball = Launch Speed = V₀ = ?

    Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)

    a = acceleration = 376 m/s²

    t = time = 0.058 s

    Therefore,

    V₀ = 0 m/s + (376 m/s²)(0.058 s)

    V₀ = 21.81 m/s

    Now, for x-component:

    V₀ₓ = V₀ Cos θ

    where,

    V₀ₓ = x-component of launch velocity = ?

    θ = Angle with horizontal = 59.9⁰

    V₀ₓ = (21.81 m/s)(Cos 59.9°)

    V₀ₓ = 10.94 m/s

    for y-component:

    V₀ₓ = V₀ Sin θ

    where,

    V₀y = y-component of launch velocity = ?

    θ = Angle with horizontal = 59.9⁰

    V₀y = (21.81 m/s)(Sin 59.9°)

    V₀y = 18.87 m/s

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