Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 weeks ago

Plz help..A baseball leaves the bat with a velocity of 55 m/s.?

The ballpark fence is 120 m away.Does the ball reach the fence if it leaves the bat traveling upward at an angle of 30 degrees to the horizontal.

Plz show work and equations/formulas.

4 Answers

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  • 2 weeks ago
    Favorite Answer

    vertical velocity Vv = 55 sin 30 = 27.5 m/s

    horizontal velocity Vh = 55 cos 30 = 47.6 m/s

    time to reach peak = Vv/g = 27.5/9.8 = 2.81 sec

    time to fall from peak, same 2.61 sec

    time in air = 5.61 sec

    horizontal distance traveled in that time = 47.6 m/s x 5.61 s = 267 m

    yes, it does reach the fence

    • Pinkgreen
      Lv 7
      2 weeks agoReport

      My opinion: it was proper to say that it actually flied over the fence.

  • 2 weeks ago

    The ball’s initial velocity has a vertical and horizontal velocity.

    Vertical = 55 * sin 30 = 27.5 m/s

    Horizontal = 55 * cos 30 = 27.5 * √3 (48 m/s)

    During the time the ball is in the air, two things are happening. It has a vertical acceleration of -9.8 m/s^2. And the horizontal velocity is constant. During one half of the total time, the ball’s vertical velocity decreases from 27.5 m/s to 0 m/s. Let’s use the following equation to calculate the time for this to happen.

    vf = vi + a * t

    0 = 27.5 + -9.8 * t

    t = 27.5 ÷ 9.8

    This is approximately 2.8 seconds. So, the total time is approximately 5.6 seconds. To calculate the maximum horizontal distance the ball moves, multiply its horizontal velocity by the total time.

    d = 27.5 * √3 * 2 * 27.5 ÷ 9.8 = 1,512.5 * √3 ÷ 9.8

    This is approximately 267 meters. Since the fence is only 120 meters from the ball park, the answer is yes, I remember a much easier way to calculate this distance.

    Range = v^2/g * sin 2θ

    Range = 55^2/9.8 * sin 60

    This is approximately 267 meters. If you use this equation, you do not need to calculate the time. I hope this is helpful for you.

  • oubaas
    Lv 7
    2 weeks ago

    range R = V^2/g*sin (2*30) = 55^2/9.806*0.866 = 267 m >> 120

  • 2 weeks ago

    x'=55cos(30*)=>x=55cos(30*)t

    y=55sin(30*)t-9.8(t^2)/2, at time t s

    where 9.8 m/s^2 is the gravitational acceleration.

    When the ball lands, y=0=>

    t(55sin(30*)-4.9t)=0=>

    t=55(0.5)/4.9

    x=55cos(30*)[55(0.5)/4.9]=>

    x=267.32 m>120 m

    Thus, the ball flies over the fence.

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