A helicopter is flying horizontally at 8.0m/s when it drops a package. a) How much time elapses before the velocity of the package doubles? >> Is the "8.0m/s" horizontal motion of the helicopter of any relevance for this question?

No, the horizontal velocity is not necessary. However, the question itself is impossible without an initial vertical velocity, for if we are to assume that the initial vertical velocity is 0 m/s, then the elapsed time before the velocity doubles is undetermined.

I disagree. Initially the package has only a horizontal component to velocity (8.0 m/s). However, immediately the package begins to free-fall. As soon as it free-falls, there will be a vertical component to the velocity in addition to the fixed horizontal motion (assuming no wind resistance is considered). When that happens, the true velocity will be the magnitude of the vector of motion, or more simply, the result of the Pythagorean Theorem on the horizontal and vertical components of the vector of motion. So, first you must determine what the vertical velocity component must be for the overall velocity to become double (ie, 16.0 m/s). That comes from Pythagorean:\[h^2+v^2=vel^2\]You know that h = 8.0 and vel = 16.0, therefore you can solve for the vertical velocity needed (v). Once you know what v should be, keep in mind that free-fall velocity is given by v = -9.8t, so you can substitute the necessary v velocity and solve for time.

It depends on when the parachute attached to the package opens.

Well, if you want to go THERE...it also depends on the shape of the object, the shape of the parachute, whether or not the person who packed the parachute was drunk that day, etc. :-)

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