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6 August, 17:08

A golf ball (m = 46.0 g) is struck with a force that makes an angle of 44.4° with the horizontal. the ball lands 193 m away on a flat fairway. if the golf club and ball are in contact for 6.92 ms, what is the average force of impact? (neglect air resistance.)

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  1. 6 August, 18:42
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    Using the formula for bodies in projectile motion, we can calculate for the velocity v:

    d = (v^2) * (sin2theta) / g

    Substituting the given values:

    193 m = (v^2) * (sin (2*44.4)) / (9.8 m/s^2)

    v = 50.5 m/s

    Calculating for the impulse I,

    I = mf * vf - mi * vi

    since mf = mi and vi = 0 (the golf ball starts form rest)

    I = m * vf

    I = 0.046 kg * 50.5 m/s

    I = 2.323 kg m/s

    Calculating for the force of impact:

    I = F t

    F = (2.323 kg m/s) / (6.92 x 10^-3 s)

    F = 335.7 kg m/s^2 = 335.7 N
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