DANE
β=30°
α=60°
V=15w/h=15*1852m/3600s=5,592m/s
t=15min=900s
d=vt=5,592*900=5932.5m
d=h*ctgβ-h*ctgα
h=d/(ctg-ctgα)=5932.5/(√3-√3/3)=5932.5/(2/3*√3)=2283,42m ??
BL=h*ctgα=2283,42*√3/3=1318,33m
ODP
BL=1,318km
Pozdrawiam
PS
zminilem plus na minus
inny rysunek punkt B po prawej
Dosc wysoka ta lampa
Hans