A baseball outfielder throws a 0.150 kg baseball at a speed of 37.0 m/s and an initial angle of 27.0°. What is?

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A baseball outfielder throws a 0.150 kg baseball at a speed of 37.0 m/s and an initial angle of 27.0°. What is the kinetic energy of the baseball at the highest point of its trajectory?

3 Answers

  • The ball is released at a starting height of 0, at an angle of 27° up from horizontal.

    At the highest point of the trajectory the vertical velocity = 0. The horizontal velocity is constant at 37cos(27°) = 33.0 m/s

    E = (1/2) m v^2

    = 0.5 x 0.150 x (33^2)

    = 81.7 J

    ———————————————————————————–

    Separately, if you are asked:

    The vertical component of the initial velocity is 37sin(27°) = 16.8 m/s. Max height = vt + 1/2 g(t^2), where g = acceleration due to gravity = -9.8 m/(s^2) and t is time

    We can find t from this relation:

    v + gt = 0, since the vertical velocity of the ball at the peak is 0.

    37 -9.8t = 0

    37 = 9.8t

    t = 0.26

    Plug into max height equation:

    =(37)(0.26) + 1/2 g(0.26^2)

    =9.3 m

  • At the highest point:-

    =>Uy = 0

    =>KE = 1/2m[Ux]^2

    =>KE = 1/2 x 0.15 x [ucosA*]^2

    =>KE = 1/2 x 0.15 x [37 x cos27*]^2

    =>KE = 81.51 J

  • the guy above me is incorrect. The y values and x values are autonomous of the different. I.e. in simple terms because of the fact the ball isnt shifting up doesnt propose its no longer shifting suited. What you are able to desire to do is locate the rate of the ball interior the x direction. I dont sense like doing it , yet use trig residences. SOH CAH TOA ….etc….

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