Ahelicopter is ascending vertically with a speed of 5.13 m/s . at a height of 112 m above the earth, a package is dropped from the helicopter. how much time does it take for the package to reach the ground? [hint: what is v0 for the package? ]
express your answer to three significant figures and include the appropriate units.

It takes the package 5.33 s to reach the ground.

Explanation:

The equation for the height of the package at a time "t" is as follows:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Initially, the package is moving at 5.13 m/s relative to the ground because it moves with the helicopter. Then, v0 = 5.13 m/s. When the package reaches the ground, its height will be 0 (placing the origin of the frame of reference on the ground). Then, we can calculate the time it takes the package to reach the ground using the equation of height of the package:

y = y0 + v0 · t + 1/2 · g · t²

0 = 112 m + 5.13 m/s · t -1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 5.13 m/s · t + 112 m

Solving the quadratic equation:

t = 5.33 s ( the other negative value is discarded because the time can´t be negative).

It takes the package 5.33 s to reach the ground.

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step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6