Observatory 71934

The aircraft flying towards the observatory was aimed at a distance of 5300 m at an elevation angle of 28º and after 9 seconds at a distance of 2400 m at an elevation angle of 50º. Calculate the distance the plane has flown in this time interval, its speed, and the change in altitude.

Correct answer:

x =  3203.5048 km
v =  355.945 m/s
h =  649.6926 m

Step-by-step explanation:

a=5300 m α=28  t=9 s  b=2400 m β=50   sin α = h1:a cos α = x1:a  h1=a sinα=a sin28° =5300 sin28° =5300 0.469472=2488.19928 m x1=a cosα=a cos28° =5300 cos28° =5300 0.882948=4679.62224 m  sin β = h2:b cos β = x2:b  h2=b sinβ=b sin50° =2400 sin50° =2400 0.766044=1838.50666 m x2=b cosβ=b cos50° =2400 cos50° =2400 0.642788=1542.69026 m  h=h1h2=2488.19931838.5067649.6926 m x=x1x2=4679.62221542.69033203.5048 m  x=h2+x2=649.69262+3203.50482 m=3203.5048 km
v=tx=93203.5048355.945 m/s  v2=v km/h=v 3.6  km/h=355.945 3.6  km/h=1281.402 km/h



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