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A batter hits a fly ball which leaves the bat 0.98 m above the ground at an angle of 60 degree with initial speed 28 m/s?

a) How far from home plate would the ball land if not caught?

(b) The ball is caught by the centerfielder who, starting at a distance of 103 m from home plate, runs straight toward home plate at a constant speed and makes the catch at ground level. Find his speed.

m/s

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  • 10 months ago
    Favorite Answer

    eTo solve this problem, we need to determine the time the baseball is in the air. In the following work, I will be doing this. The first steps are to determine the vertical and horizontal components of the baseball’s initial velocity.

    Vertical = 28 * sin 60 = 14 * √3 m/s

    Horizontal = 28 * cos 60 = 14 m/s

    During the time the baseball is in the air, it has a vertical acceleration of -9.8 m/s^2. Let’s use the following equation to determine the baseball’s vertical velocity the moment it lands.

    4 * -9.8 * -0.98

    vf^2 = vi^2 + 2 * a * d

    a = -9.8 m/s^2, d = final height – initial height = 0.0 – 0.98 = -0.98 meter

    vf^2 = 588 – 2 * -9.8 * -0.98

    vf^2 = 588 – 192.08

    vf = ±√568.792

    This is approximately 23.8 m/s. But since the baseball is falling, the vertical velocity is negative. Let’s use the following equation.

    vf = vi + a * t, a = -9.8 m/s^1

    -√568.792 = 14 * √3 – 9.8 * t

    t = [(-√568.792 – 14 * √3)] ÷ 9.8

    The time is approximately 4.9 seconds.

    a) How far from home plate would the ball land if not caught?

    d = 14 * [(-√568.792 – 14 * √3)] ÷ 9.8

    This is approximately 69.7 meters. I am still working.

    (b) The ball is caught by the centerfielder who, starting at a distance of 103 m from home plate, runs straight toward home plate at a constant speed and makes the catch at ground level. Find his speed.

    To answer this problem, I must assume that he starting running the second the baseball was hit.

    Speed d ÷ = 69.7 ÷ *([(-√568.792 – 14 * √3)] ÷ 9.8

    This is approximately 14.3 seconds. I hope this is helpful for you.

    • ...Show all comments
    • Whome
      Lv 7
      9 months agoReport

      You should thank her. It is certainly not deserved. You claim more precision than the question numbers might suggest possible and completely missing the answer on the second part.You give a speed in units of seconds while the cardinal should be the distance traveled over time...33 m / 5 s

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  • Whome
    Lv 7
    10 months ago

    Use the general equation for position in time with constant acceleration.

    s = s + v₀t + ½at²

    In the vertical, with up positive and the ground the origin, the equation would be

    0 = 0.98 + (28sin60)t + ½(-9.8)t²

    0 = 0.98 + 24.25t - 4.905t²

    quadratic formula

    t = (-24.25 ±√(24.25² - 4(-4.905)(0.98))) / (2(-4.905))

    t = - 0.040 s ignore

    or

    t = 4.989... s is the time the ball is in the air.

    in that time, assuming home base is origin and towards outfield is positive, the ball travels horizontally

    s = 0 + (28sin60)(4.989) + ½(0)(4.989²)

    s = 69.8489...

    s = 70 m ANSWER a

    Assuming the fielder starts running the moment the ball is hit,

    v = d/t

    v = (103 - 70) / 4.989

    v = 6.6445...

    v = 6.6 m/s

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  • 10 months ago

    R = (v₀²/2g)(1 + √(1 + (2gy₀/v₀²sin²θ))) sin(2θ)

    R = (28²/19.6)(1 + √(1 + (19.6•0.98/28²sin²60))) sin(120)

    R = (40)(1 + √(1 + (19.2/784•0.75))) 0.866

    R = (40)(1 + √(1.0326531)) 0.866

    R = (40)(1 + 1.016195) 0.866

    R = 69.8 m

    distance he runs is 103– 69.8 = 33.2 m

    but we don't know his starting time...

    range of object projected at an angle θ

    R = (V²/g)sin(2θ)

    V is initial velocity in m/s

    g is the acceleration of gravity 9.8 m/s²

    Time of flight = R/(Vcosθ)

    Max height

    h = (V²/2g)sin²θ

    R/h = cotθ

    different heights:

    R = (v₀²/2g)(1 + √(1 + (2gy₀/v₀²sin²θ))) sin(2θ)

    d is the total horizontal distance travelled by the projectile.

    v is the velocity at which the projectile is launched

    g is the gravitational acceleration—usually taken to be

    9.81 m/s² near the Earth's surface

    θ is the angle at which the projectile is launched

    y₀ is the initial height of the projectile

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