Answer to Question #299714 in Mechanics | Relativity for Darling Pusta

Question #299714

Given, Sketch, And Solution of the problem?


1
Expert's answer
2022-02-21T12:11:13-0500

Answer

Range of object projected at an angle θ

"R=\\frac{v^2\\sin2\\theta}{g}"


Now putting angle of 30° and 50°

So range becomes

"R=\\frac{400^2\\sin2\\times30\u00b0}{9.81}=14140m"


And

"R=\\frac{400^2\\sin2\\times50\u00b0}{9.81}=16080m"






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