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A ball is dropped from 15.6m. Another ball is thrown upwards. The motions are reverse of each other. How to find the location they cross paths?
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let the height h=15.6m let the height of cross point frm ground is =x so the falling ball will travell 15.6-x now we will use the equation h=ut+(1/2)gt^2 this equation is for free fall ball will b 15.6-x=ut+(1/2)gt^2 15.6-x=0*t+(1/2)gt^2.............(1) for upward going ball this equation will b x=ut-(1/2)gt^2.................(2) now u will add equation 1 and rquation 2
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