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 equationx(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 becomesx(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 = 0Given that the height of storey 2 is 58.4 ft., formulate the quadratic equation giving height as a function of time t, for an object dropped from storey 2. Determine the solutions for this equation. Again, interpret your solutions – how long does it take for an object dropped from storey 2 of the Leaning Tower to hit the ground?Well, the initial height, x(t=0) of the object dropped from storey 2 of the Tower is 58.4ft. Theres no initial velocity, so v(t=0)=0, and we want to know at which value of t the height x(t) of the object above the ground is zero (the object is on the ground). So we have an equation:
x(t)=0=-16t^2 + 58.4
-16t^2 + 58.4 = 0
-16t^2 = -58.4
t^2 = (-58.4)/(-16) = 3.65
t is the square root of 3.65 or t is the negative square root of 3.65. Since the negative value of t makes no sense in the problem, we conclude that t=sqrt(3.65)=1.9104973 or 1.91 seconds.

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