An aeroplane takes off at an angle of 60 degree to the horizontal if the components of the velocity along the horizontal is 200 m/s what is the actual velocity? Also find the vertical components of its velocity
vx=v⋅cos60°→v=vx/cos60°=200/cos60°=v_x=v\cdot\cos60°\to v=v_x/\cos60°=200/\cos60°=vx=v⋅cos60°→v=vx/cos60°=200/cos60°=
=400 (m/s)=400\ (m/s)=400 (m/s)
vy=v⋅sin60°=400⋅sin60°=346 (m/s)v_y=v\cdot\sin60°=400\cdot\sin60°=346\ (m/s)vy=v⋅sin60°=400⋅sin60°=346 (m/s)
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