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A batsman hits a cricket ball 'off his toes' towards a fieldsman who is 65 m away. The ball reaches a maximum height of 4.9 m and the horizontal compoenent of its velocity is 28 m/s. Find the constant speed with which the fieldsman must run forward, starting at the instant the ball is hit, in order to catch the ball at a height of 1.3m above the ground. (Use g = 9.8)

Thanks in advance :)

  • math -

    At the maximum height of 4.9m, the ball has risen for 1 sec. (That's the amount of time it takes to fall from 1m height to 0.)

    So, v = 9.8-9.8t
    h = -4.9t^2 + 9.8t since h(0) = 0
    at t=1.857 sec, h = 1.3, x=52 m

    So, the fielder must run 65-52 = 13m in 1.857sec, or 7m/s

    (assuming he doesn't extend his arm too far to grab the ball!)

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