projectile question?

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A golfer hits a ball from the origin with an initial speed of 15.0 m/s at an angle of 62.0 degrees above the horizontal. The ball is in the air 2.58 s, and lands on a green that is above the level where the ball was struck.What is the balls speed just before it lands? give me in detail.

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  • JOHN
    Lv 7
    5 months ago

    For the vertical component of the velocity just before ball lands, use v = u + at: v = 15sin62° - 9.8 x 2.58.

    Horizontal component of the velocity just before ball lands = 15cos62°. Speed on landing is

    √[(15sin62° - 9.8 x 2.58)² + (15cos62)²] = 13.9m/s.

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  • 5 months ago

    This is not difficult as long as you have memorized the relations.

    Vx(t) = Vx(0) =15cos62 = 7.042m/s ignoring air resistance

    Vy(t) = Vy(0) - g*t = 15sin62 - 9.8*2.58s = -12.04m/s

    Vf² = 7.042²+12.040 = 13.95²

    • Lôn
      Lv 7
      5 months agoReport

      Vertical speed = 15sin62.... 62 is the angle to the horizontal, not the vertical.

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  • 5 months ago

    How long (in seconds) does it take for the coin to travel 64 m upward from the point it was launched

    Since the coin is moving vertically, it has an acceleration of -9.8 m/s^2. Let’s use the following equation to determine the time.

    d = vi * t + ½ * a * t^2

    64 = 37 * t + ½ * -9.8 * t^2

    4.9 * t^2 – 37 * t + 64 – 0

    t = [37 ± √(-37^2 – 4 * 4.9 * 64)] ÷ 9.8

    t = [37 ± √114.6] ÷ 9.8

    t = [37 + √114.6] ÷ 9.8

    Rounded to two significant digits, the time is 4.9 seconds.

    OR

    t = [37 – √114.6] ÷ 9.8

    Rounded to two significant digits, the time is 2.7 seconds. The longer time is when the coin is falling. To prove that this is true, let’s use the following equation to determine the coin’s velocity at this time.

    vf = vi + a * t

    vf = 37 + -9.8 * 4.9 = -11.02 m/s

    Since the velocity is negative, the coin is falling. I hope this is helpful for you.

    Since the coin is moving vertically, it has an acceleration of -9.8 m/s^2. Let’s use the following equation to determine the time.

    d = vi * t + ½ * a * t^2

    64 = 37 * t + ½ * -9.8 * t^2

    4.9 * t^2 – 37 * t + 64 – 0

    t = [37 ± √(-37^2 – 4 * 4.9 * 64)] ÷ 9.8

    t = [37 ± √114.6] ÷ 9.8

    t = [37 + √114.6] ÷ 9.8

    Rounded to two significant digits, the time is 4.9 seconds.

    OR

    t = [37 – √114.6] ÷ 9.8

    Rounded to two significant digits, the time is 2.7 seconds. The longer time is when the coin is falling. To prove that this is true, let’s use the following equation to determine the coin’s velocity at this time.

    vf = vi + a * t

    vf = 37 + -9.8 * 4.9 = -11.02 m/s

    Since the velocity is negative, the coin is falling. I hope this is helpful for you. I use the following website to check my answers for quadratic equations. This is how I can know if my answers are correct.

    http://www.math.com/students/calculators/source/qu...

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