A package is dropped from a helicopter moving upward at 15 m/s. If it takes 8.0 s before it hits the ground, how high above the ground was the package when it was released?

Answer:

Kinematics problem.

We will use the equation y = y₀ + v₀t + 1/2at²

Where y is the final height (0m, or the ground), y₀ is the initial height (what we are solving for), v₀ is the initial velocity (15m/s), a is acceleration (-9.8 due to gravity) and t is the time (8.0s)

y = y₀ + v₀t + 1/2at²

0 = y₀ + 15(8) + 1/2(-9.8)(8)²

0 = y₀ + 120 - 313.6

0 = y₀ - 193.6

y₀ = 193.6m, this is how high above the ground the package was.