A golfer hits a shot to a green that is elevated 2.80 m above the point where the ball is struck. The ball leaves the club at a speed of 18.

Question

A golfer hits a shot to a green that is elevated 2.80 m above the point where the ball is struck. The ball leaves the club at a speed of 18.7 m/s at an angle of 49.0˚ above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

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Thiên Hương 4 years 2021-08-28T07:19:40+00:00 1 Answers 26 views 0

Answers ( )

  1. binhan
    0
    2021-08-28T07:21:15+00:00
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    Answer:

    32.812m/s

    Explanation:

    Now the time of the projectile motion is given by;

    t = usinA/ g

    Where A is angle =49°

    u is initial velocity,u = 18.7m/s

    Hence t = 18.7 ×sin49°/ 9.8 = 1.44s

    The final velocity from Newton’s law V = U + gt

    = 18.7 + (9.8 ×1.44)= 32.812m/s

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