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anonymous

  • 5 years ago

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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  1. anonymous
    • 5 years ago
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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

  2. anonymous
    • 5 years ago
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    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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