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Algebra

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

Well, we trudge up to a height of 77.7 ft. At this point, Galileo is exhausted. Consequently, in frustration on the third storey, he throws an object toward the earth (rather than just dropping it) at a rate of 20 ft / sec. Thus, on storey 3, at a height of 77.7 ft, our object has an initial velocity of 20 ft / sec. downward.
Formulate the quadratic equation, giving height as a function of time, for our object thrown from the 3rd storey of the Leaning Tower. Again, determine the solutions for this equation and interpret the solutions. How long does it take for our object, thrown off of storey 3, to hit bottom?

Solution Preview

The equation is x(t) = -16t^2 + 20t + 77.7

-16t^2 + 20t + 77.7 = 0

a = -16, b ...

Solution Summary

A complete, neat and step-by-step solution to the algebra question regarding boulders being dropped is provided.

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