Direction of a plane proof 48. A pilot wants to land at a small lake that is [N30.0 ̊W] of the airport that she is starting from. The wind has a velocity of 25.0 m/s[W] and the air speed of the plane is 1.90 × 10^2 m/s . What direction will the plane have to fly to get to its destination? What will be the velocity of the plane relative to the ground? vt as the speed you'll achieve . x as the plane speed or velocity. w as wind speed or velocity. Θ as the angle the plane has to go to achieve angle Φ at vt speed. Φ as the angle away from start. Know start to take the big picture coloured in red vt^2=(xsinΘ+w)^2+(xcosΘ)^2 vt^2=(x^2sinΘ^2+wxsinΘ+wxsinΘ+w^2)+(x^2cosΘ^2) now we have to get rid of the cosΘ^2 by taking(in green) x^2=(x^2sinΘ^2)+(x^2cosΘ^2) With x^2 divided x^2/x^2=(x^2sinΘ^2)/x^2+(x^2cosΘ^2)/x^2 1=sinΘ^2+cosΘ^2 1-cosΘ^2=sinΘ^2 Now sub in to the red coloured equation vt^2=x^2(1-cosΘ^2)+wxsinΘ+wxsinΘ+w^2)+(x^2cosΘ^2) vt^2=(x^2-x^2cosΘ^2)+(wxsinΘ+wxsinΘ+w^2)+(x^2cosΘ^2) vt^2=x^2+2wxsinΘ^2+w^2 Now look at what's coloured in blue which is used to get rid of vt sinΦ=(xsinΘ+w)/vt (vt=(xsinΘ+w)/sinΦ)^2 vt^2=(x^2sinΘ^2+wxsinΘ+wxsinΘ+w^2)/sinΦ^2 Now we sub this equation into the red and green equation (x^2sinΘ^2+2wxsinΘ+w^2)/sinΦ^2=x^2+2wxsinΘ+w^2 Move everything over to left (x^2sinΘ^2/sinΦ^2)+(2wxsinΘ/sinΦ^2-2wxsinΘ) (+w^2sinΦ^2-x^2-w^2)=0 A B C Use the quadratic formula sinΘ=-(2wx/sinΦ^2-2wx)+/-square root((2wx/sinΦ^2-2wx)^2-4x^2/sinΦ^2(w^2/sinΦ^2-x^2-w^2)/2x^2/sinΦ^2 sinΘ=(-2wx/sinΦ^2+2wx)+/-square root((4x^2w^2/sinΦ^4-4w^2x^2/sinΦ^2-4w^2x^2/sinΦ^2+4w^x^2)(-4x^2w^2/sinΦ^4+4x^4/sinΦ^2+4w^2x^2/sinΦ^2))/2x^2/sinΦ^2 sinΘ=(-2wx/sinΦ^2+2wx)+/-square root((-4w^2x^2/sinΦ^2+4w^x^2+4x^4/sinΦ^2))/2x^2/sinΦ^2 sinΘ=(-2wx/sinΦ^2+2wx)+/-square root((-4w^2x^2+4x^4)/sinΦ^2+4w^2x^2)/2x^2/sinΦ^2 multiply the whole equation by sinΦ^2 sinΘ=(-2wx+2wxsinΦ^2)+/-square root((-4w^2x^2+4x^4)sinΦ^2+4w^2x^2sinΦ^4)/2x^2 divid the equation by 2 sinΘ=(-wx+wxsinΦ^2)+/- square root((-w^2x^2+x^4)sinΦ^2+w^2x^2sinΦ^4)/x^2 Divid by x sinΘ=(-w+wsinΦ^2)+/-square root((-w^2+x^2)sinΦ^2+w^2sinΦ^4)/x now say + not +/- and factor out sinΦ out of what's in the Square root since we don't want a negative angle. sinΘ=(-w+wsinΦ^2)+square root((-w^2+x^2+w^2sinΦ^2))sinΦ/x Also written as sinΘ=square root((-w^2+x^2+w^2sinΦ^2))sinΦ+(-w+wsinΦ^2)/x if wind is in y direction cosΘ=square root((-w^2+x^2+w^2cosΦ^2))cosΦ+(-w+wcosΦ^2)/x x>w If wind is going to opposite your going make negative.