question archive A package is dropped from a helicopter moving upward at 15 m/s

A package is dropped from a helicopter moving upward at 15 m/s

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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?

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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.