A plane covers 500 ,m in a direction 60^o wit | vedclass

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Question

(A) Let the total distance covered be $d = 500 \,m$ at an angle $	heta = 60^o$ with the horizontal runway.The horizontal component of the distance is

Vedclass · Accepted Answer

Horizontal: $250 \,m$,Vertical: $250 \sqrt{3} \,m$

Vedclass · Answer

(A) Let the total distance covered be $d = 500 \,m$ at an angle $	heta = 60^o$ with the horizontal runway.The horizontal component of the distance is given by $d_x = d \cos(	heta)$.$d_x = 500 \cos(60^o) = 500 	imes (1/2) = 250 \,m$.The vertical component of the distance is given by $d_y = d \sin(	heta)$.$d_y = 500 \sin(60^o) = 500 	imes (\sqrt{3}/2) = 250 \sqrt{3} \,m$.Therefore,the horizontal distance is $250 \,m$ and the vertical distance is $250 \sqrt{3} \,m$.

$A$ plane covers $500 \,m$ in a direction $60^o$ with the runway while starting the flight. Find the distance covered by the plane in the horizontal and vertical directions.

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