A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was realeased?

Answer 1

The time to V = 0 derives from Vf = Vo - gt or 0 = 15 - 9.8t yielding t(up) = 1.53sec. resulting in t(dwn) = 16 - 1.53 = 14.47 sec.

h(dwn) then derives from h = Vo(t) + g(t^2)/2 or h = 0 + 4.9t(dwn)^2 yielding h(dwn) = 4.9(14.47)^2 = 1025.96 met.

h(up) then derives from h(up) = 15(1.53) - 4.9(14.47)^2 = 11.48 met.

The vertical rise from ejection to zero vertical velocity derives from h(up) = 15(1.53) - 4.9(1.53)^2 = 11.48 met.

Therefore, the altitude of release is h(rel) = 1025.96 - 11.48 = 1014.48 met.

Answer 2

gasagwa