Physics

Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the ground. Upon contact with the bat, the ball is 1 m above the ground. Payer B wishes to duplicate this bunt, in so far as he also wants to give the ball a velocity parallel to the ground and have his ball travel the same horizontal distance as player A's ball does. However, player B hits the ball when it is 1.8 m above the ground. What is the magnitude of the initial velocity that player B's ball must be given?

Does anyone know how to do this problem? I am very confused

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asked by Jerry
  1. well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

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  2. First you need to solve for time by using

    d=(1/2)(a)(t^2)+(vi)t
    1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s
    t=.45 sec

    Then you find the horizontal distance traveled by using

    v=d/t
    1.3m/s=d/.54sec
    d=.585m

    Then you need to find the time of player B by using

    d=(1/2)(a)(t^2)+(vi)t
    1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0
    t=.61 sec

    Finally to find player Bs initial horizontal velocity you use the horizontal equation

    v=d/t
    v=.585m/.61 sec

    so v=.959m/s

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    posted by rachel

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