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13 June, 06:38

A football is thrown toward a receiver with an initial speed of 15.4 m/s at an angle of 36.2 ◦ above the horizontal. At that instant, the receiver is 15.8 m from the quarterback. The acceleration of gravity is 9.81 m/s 2. With what constant speed should the receiver run to catch the football at the level at which it was thrown? Answer in units of m/s

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  1. 13 June, 10:17
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    vr = 3.906 m/s

    Explanation:

    vi = 15.4 m/s

    ∅ = 36.2º

    d = 15.8 m

    g = 9.81 m/s²

    vr = ?

    We have to get Xmax = R as follows

    R = vi²*Sin (2∅) / g

    ⇒ R = (15.4 m/s) ²*Sin (2*36.2º) / (9.81 m/s²)

    ⇒ R = 23.0437 m

    then we can get t applying the equation

    R = vi*Cos ∅*t ⇒ t = R / (vi*Cos ∅)

    ⇒ t = (23.0437 m) / (15.4 m/s*Cos 36.2º) = 1.85 s

    Now, we find the distance that the receiver must be run

    x = R - d

    ⇒ x = 23.0437 m - 15.8 m = 7.2437 m

    Finally, we get the speed

    vr = x / t

    ⇒ vr = 7.2437 m / 1.85 s = 3.906 m/s
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