Question

A package is dropped from a helicopter that is moving upward at 20.0 m/s.

If it takes 18.0s before the package strikes the ground, how high above the ground was the package when it was released?

Answer 1

Step-by-step explanation:

We have for velocity of package v= 20 - 9.8t and for distance from the ground y = h + 20t - 4.9t2.

We have y = 0 and t = 18.0 s we have h = 4.9·182 - 20·18 = 1227.6 m.

Velocity at landing v = 20 - 9.8·18 = - 156.4 m/s