Question

68. A body is projected such that its K.E. at the top is th of its initial.RE. What is the angle of projection with the horizontal? (a) 30° (b) 45° (d) 90° (c) 600 nn 11 ooted with

NEET/Medical Exams

Physics

Solution

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ut imtial wetouty = 4 And angle of ucloity with houicontals Sinu the westial neloily is oat top pomt and hovizontal uclority doesin't change, the ke at the ( quad ) Lop ( quad=frac{1}{2} m u^{2} cos ^{2} x ) and initial ( k epsilon=frac{1}{2} m u^{2} ) Ginen that KE at top ( =frac{3}{4} ) of initial ( k in ) [ begin{array}{c} frac{1}{2} y^{2}-cos ^{2} x=frac{3}{4} x frac{1}{x} operatorname{mos}^{2} cos ^{2} x=frac{3}{4} cos x=sqrt{frac{3}{4}}=frac{sqrt{3}}{2} Rightarrow x=30^{circ} therefore A u=(a) cdot 30^{circ} end{array} ]