A man hits a 50 grams golf ball such that it leaves the tee at an angle of 40o with the horizontal and strikes the ground at the same elevat

Question

A man hits a 50 grams golf ball such that it leaves the tee at an angle of 40o with the horizontal and strikes the ground at the same elevation 20 m away. Determine the impulse of the club C on the ball.

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Diễm Thu 4 years 2021-08-12T23:22:23+00:00 1 Answers 20 views 0

Answers ( )

  1. Tryphena
    0
    2021-08-12T23:23:47+00:00
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    Answer:

    Explanation:

    Range of projectile R = 20 m

    formula of range

    R = u² sin2θ / g

    u is initial velocity , θ is angle of projectile

    putting the values

    20 = u² sin2x 40 / 9.8

    u² = 199

    u = 14.10 m /s

    At the initial point

    vertical component of u

    = u sin40 = 14.1 x sin 40

    = 9.06 m/s

    Horizontal component

    = u cos 30

    At the final point where the ball strikes the ground after falling , its speed remains the same as that in the beginning .

    Horizontal component of velocity

    u cos 30

    Vertical component

    = – u sin 30

    = – 9.06 m /s

    So its horizontal component remains unchanged .

    change in vertical component = 9.06 – ( – 9.06 )

    = 18.12 m /s

    change in momentum

    mass x change in velocity

    = .050 x 18.12

    = .906 N.s

    Impulse = change in momentum

    = .906 N.s .

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