A cannon fires an acrobat into the air at an angle of 45 above the horizontal such that she reaches a height h above her launch height. If the cannon is now aimed straight upwards, what is the maximum height reached by the same acrobat

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A cannon fires an acrobat into the air at an angle of 45 above the horizontal such that she reaches a height h above her launch height. If the cannon is now aimed straight upwards, what is the maximum height reached by the same acrobat

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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first u need to know the formula h(maximum ht above ground)=(usin@)^2/2g so apply that with g=10m/s^2 h and angle 45 then u get initial velocity u in terms of h when they say u aim verticallly it means straight up with 90 degree angle with vertical take v=0 use v^2-u^2=2as find s ht it reaches u have any doubts on this problem u can ask me
oh man... accidently deleted the question about the speed trap, real fast, how do you solve for t? 10t = t^2?

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