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 co

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 1.2 m above the ground. Player 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 1.6 m above the ground. What is the magnitude of the initial velocity that player B’s ball must be given

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  1. Answer:

    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

    Explanation:

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  2. Answer:

    Explanation:

    cSep 20, 2010

    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.

    Anonymous

    Sep 20, 2010

    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

    Reply

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