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3 July, 02:19

A football is thrown toward a receiver with an initial speed of 16.9 m/sat an angle of 36.5◦ above the horizontal. At that instant, the receiver is 16.8 m from the quarterback. The acceleration of gravity is 9.81 m/s2. 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. 3 July, 02:27
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    vr = 5.39 m/s

    Explanation:

    vi = 16.9 m/s

    ∅ = 36.5º

    d = 16.8 m

    g = 9.81 m/s²

    vr = ?

    We have to get Xmax = R as follows

    R = vi²*Sin (2∅) / g

    ⇒ R = (16.9 m/s) ²*Sin (2*36.5º) / (9.81 m/s²)

    ⇒ R = 27.842 m

    Now we can get t, using the formula:

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

    ⇒ t = (27.842 m) / (16.9 m/s*Cos 36.5º) = 2.049 s

    We get x (the distance which the receiver must be run) as follows

    x = R - d

    ⇒ x = 27.842 m - 16.8 m = 11.042 m

    Finally we have

    vr = x / t

    ⇒ vr = 11.042 m / 2.049 s = 5.39 m/s
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