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baseball is hit with a speed of 30.0 m/s at an angle of 50 It lands on the flat roof of a 10m tall nearby building, If the ball was hit when it was 1.5 m above the ground, what horizontal distance does it travel before it lands on the building?

  • physical -

    V 0x = 30*cos 50 = 19.28 m/s
    V 0y = 30*sine 50 = 22.98

    Y = 14-1.5 =12.5
    Y = V 0y*t -0.5*g*t^2
    12.5 = 22.98*t -4.9*t^2
    Solve the quadratic equation for t. t = 4.06 seconds

    X = v 0x*t
    X = 19.28*4.06 = 78.27 meters

  • physical -

    The ball travels in the vertical direction with initial vertical component of 30 sin 50 [22.98 m/s^2] upward till its velocity component in vertical direction becomes let us say V. The final vertical component after travelling [14 - 1.5] = 12.5 m is given by
    [30 sin 50]^2 - V^2 = 2x9.8x12.5 or

    V^2 = 528.14 - 245 = 283.14 or V = 16.826 m/s

    Time required to reduce the vertical component from 22.98 to 16.82 would be 6.16/9.8 s = 0.628 s

    The horizontal distance travelled by the ball in this time is

    30 cos 50 x 0.628 = 12.11 m

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