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a boy throws a ball in air in such a manner that when the ball is in its max height he throws another ball. if the balls are thrown after the time difference of 1 s ; then what will be the height attain by them:
u=0 m/st=1 secondtime of ascend,t=u/g1=u/9.89.8 m/s\maximum height reached, h=u^2/2gh=(9.8)^2/2×9.8=4.9 m
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