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dropped object time

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All boulders and rocks (large enough not to be buffeted by air current) appear to take the same time to hit bottom. My conjecture, observes Galileo, leader of the team, is that there is a universal gravitational constant which affects all objects, large or small. Knowing, as he does, that if such a constant exists, call it g for gravity, it can be plugged into our quadratic equation

x(t) = -½gt2 + v(t=0)t + x(t=0)

where v(t = 0) is our initial velocity (which is 0 if the object is dropped, rather than thrown, off of the leaning tower) and x(t = 0) is the distance that we are standing above ground level (the height at that level of the leaning tower). Well, what we know, and what Galileo is about to find out, is that our gravitational constant g = 32 ft / sec / sec. Let us say, then, that we begin by dropping rocks and boulders of various sizes off of the leaning tower. Just dropping our object, our v(t = 0) term would be 0, so that our equation becomes

x(t) = -½gt2 + x(t=0)

We have, as we know, g = 32 ft / sec / sec and the following data
storey 1 height = 39.37 ft. Consequently, our equation for storey 1 is
-16t2 + 39.37 = 0

Determine the solutions of the above quadratic equation. Interpret the solutions - how long does it take for an object dropped from storey 1 of the Leaning Tower to hit the ground?

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Solution Summary

Using square roots, dropped object time is assessed.

Solution Preview

Solving -16t^2 + 39.7 = 0, we get
-16t^2 = -39.7

Dividing both sides by ...

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