Dock diving is a great form of athletic competition for dogs of all shapes and sizes. Sheba, the American Pit Bull Terrier, runs and jumps o

Question

Dock diving is a great form of athletic competition for dogs of all shapes and sizes. Sheba, the American Pit Bull Terrier, runs and jumps off the dock with an initial speed of 8.62 m/s at an angle of 28° with respect to the surface of the water. (Assume that the +x axis is in the direction of the run and the +y axis is up.) (a) If Sheba begins at a height of 0.85 m above the surface of the water, through what horizontal distance does she travel before hitting the surface of the water? m

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Adela 5 months 2021-09-02T12:57:01+00:00 2 Answers 2 views 0

Answers ( )

  1. thugiang
    0
    2021-09-02T12:58:29+00:00
    Reply

    Answer:

    3.139 m

    Explanation:

    Velocity = 8.62 m/s

    Angle of flight = 28°

    Height above water = 0.85 m

    we are to take the +ve x-axis in the direction of run, and +ve y-axis up.

    This is a projectile situation, and the horizontal distance of travel (range) is calculated as

    R = (v^{2}sin2∅)/2g

    where g is the acceleration due to gravity = +9.81 m/s^{2} sine it acts downwards

    R = (8.62^{2}sin (2×28°))/(2 x 9.81)

    R = (74.30 x 0.8290)/19.62

    R = 61.60/19.62 = 3.139 m

  2. sori
    0
    2021-09-02T12:58:32+00:00
    Reply

    Answer:

    62.4m

    Explanation:

    a) Horizontal distance traveled = x = v_x * t

    where,

    v_x  = horizontal velocity

    and t = time in the air)

    Time in the air t can be solved using the equation for y:

    i_y=y_o+v_o_yt - 0.5gt^2

    where, y_o = initial height i.e 0.85 m

    v_o_y = initial vertical velocity

    and g = 9.8 m/s^2

    When y = 0, the dog has hit the water.

    So set y = 0 and solve for t.  

    v_o_y = 8.62 m/s \times (sin 28) = 4.04 m/s

    0 = 0.85 m + (4.04 m/s)t – 0.5gt²  

    0.5t²-4.04t-0.85  

    Solve this quadratic formula for t: the solutions are t = 8.2 s and t = -0.2. Reject the negative solution, so t = 8.2 s.

    How far does the dog travel horizontally in 9.2 s?

    x = v_x * t  = (8.62 m/s) \times (cos 28) \times  8.2 s = 62.4 m  

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