A skydiver is jumping from an airplane traveling at 10.0 m/s. The plane is 3520 m above the earth. The sky diver pulls his cord at 1760 m above the earth. Neglecting air resistance, how far does the skydiver travel horizontally before pulling the cord?

Answer 1

Δh=H-h= gt²/2
t=sqrt(2•Δh/g)=
=sqrt{2• (3520-1760)/9.8}=18.95 s.
s=vt=10•18.95=189.5 m