question archive 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?
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.