Physics

You serve a tennis ball from a height of 2.1 m above the ground. The ball leaves your racket with a speed of 19.2 m/s at an angle of 5.3° above the horizontal. The horizontal distance from the court's baseline to the net is 11.83 m, and the net is 1.07 m high. Neglect spin imparted on the ball as well as air resistance effects. Does the ball clear the net (= positive answer)? If yes, by how much? If not, by how much did it miss? In that case the answer will be negative. Answer in units of m

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asked by Derek
  1. Vo = 19.2m/s[5.3o]
    Xo = 19.2*cos5.3 = 19.11 m/s.
    Yo = 19.2*sin5.3 = 1.77 m/s.

    Tr=-Yo/g = -1.77/-9.8 = 0.181 s. = Rise
    time.

    h=ho + -Yo^2/2g = 2.1 + -(1.77^2/-19.6 = 2.1 + 0.160 m. = 2.26 m. = Max. ht. above gnd.

    d = 0.5g*t^2 = 2.26 - 1.07 = 1.19 m.
    4.9t^2 = 1.19
    t^2 = 0.243
    Tf = 0.493 s. = Fall time.

    Dx = Xo*(Tr+Tf) = 19.11 * (0.181+0.493)=
    12.88 m.

    So the ball clears the net by:
    12.88 - 11.83 = 1.05 m

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    posted by Henry

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