Answer:
Solution: a is AM⊥BF of point m, and f is FN⊥AB of point n,
AD = 24cm, then BF=24cm,
∴bn=√(bf^2+fn^2)=√(25^2-24^2)=7(cm),
∠∠AMB =∠FNB = 90, ∠ Anti-missile =∠FBN,
∴△BNF∽△BMA,
∴AB/BF=FN/AM,
∴80/25=AM/24,
Then: AM=( 16×24)/5=(384/5),
So the distance from point A to the ground is: (384/5)+4 = 404/5 (m).
So the answer is 404/5.