physics

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) A batter hits a baseball so that it leaves the bat at speed
Vo = 37.0m/s at an angle ao = 53.1°, at a location where g =
9.80 m/S2. (a) Find the position of the ball, and the magnitude and direction of its velocity, at t = 2.00 s. (b) Find the time when the ball reaches the highest point of its flight and find its height h at this point. (c) Find the Iwrizontal range R-that is, the horizontal distance from the starting point to where the ball hits the ground.

  • physics -

    (a) x = 37.0 cos53.1 * t = 22.22 t
    y = 37.0 sin53.1 t - (g/2) t^2
    = 29.59 t - 4.90 t^2
    Vx = 22.22
    Vy = 29.59 - 9.8 t
    angle to horizontal = arctan Vy/Vx
    Plug in t = 2 s

    (b) Compute y at time when Vy = 0
    t = Vyo/g = 3.02 s
    y = (Vyo)^2/(2g)

    (c) R = 2 Vo^2 sin53.1*cos53.1/g
    = 0.96 Vo^2/g

    Vo is the launch velocity and Vyo is the y component of the launch velocity

  • physics -

    24.23m/s, 23.21 degree

  • physics -

    Vo = 37 m/s
    Theta = 53.1
    Xo = 0 m i
    Yo = 0 m j
    ax = 0 m/s^2 i
    ay = -9.8 m/s^2 j

    Vox = Vo * costheta
    = 37[cos(53.1)]
    = +22.2 m/s i

    Voy = Vo * sintheta
    = 37[sin(53.1)]
    Voy = +29.6 m/s j

    x1 = +44.4 m i

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