Question

A target is fired on the top of the towed pole 13 m high. A person I standing a distance of som from the pole is capable of project ay a stone with velocity lopte ms. If he wants to Strike the target in shortest possible time, at what angle should he projecte the stone

NEET/Medical Exams

Physics

Solution

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The faster the stone is thrown it quicker it reaches the target. So let us say: ( u=10 sqrt{g} quad x=50 m quad y=13 m quad Phi= ) angle ( w r t ) horizontal. ( x=u cos phi t quad Rightarrow t=50 sec Phi / u ) ( y=u sin Phi t-g t^{2} / 2 ) ( 13=50 ) tan ( Phi-12.5 sec ^{2} Phi ) ( Rightarrow 12.5 tan ^{2} Phi-50 ) tan ( Phi+25.5=0 ) ( Rightarrow tan ^{2} Phi-4 tan Phi+2.04=0 ) ( Rightarrow tan Phi=2+-1.4 ) ( Rightarrow tan Phi=3.4 ) or 0.6 ( Rightarrow Phi=73.6^{circ} ) or ( 30.96^{circ} ) That is the answer: ( 30.96^{circ} ) Angle made by the straight line connecting the target and the man ( =tan ^{-1}(13 / 50)=14.5^{circ} ) ( left.text { time taken }left.=50^{*} sqrt{(} 1+3.4^{2}right) /(10 sqrt{mathrm{g}})=5 sqrt{(} 12.56 / mathrm{g}right) ) sec