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4 September, 17:31

Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose out line is 6.40 m from the net?

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  1. 4 September, 20:58
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    6.11°

    yes

    Explanation:

    v = 170 km/h = 47.2 m/s

    vx = v cos θ

    vy = v sin θ

    In horizontal direction, there is no acceleration.

    vx t = 11.9

    In vertical direction:

    2.5 - 0.91 = vy t + 0.5 gt²

    ⇒1.59 = 11.9 tan θ + 4.9*0.25² * sec²θ (∵g = 9.8 m/s²)

    ⇒1.59 = 11.9 tan θ + 4.9*0.25² * (1+tan²θ)

    Solving for θ will give the result 6.11°

    Displacement in x-direction

    t = 0.375 s

    vx t = 11.9

    47.2 cos (6.11°) 0.375 s = 17.6 - 11.9 = 5.7 m

    The outline is 6.40 m from the net. Thus, the ball will fall in the service box.
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