Question

the golden gate bridge in san francisco, california, was under construction from 1933-1937. suppose a construction worker dropped a tool from a point on the bridge that is 220 feet above the water. how many seconds would the tool take to hit the water? let the final height = 0 to indicate the water/ground. \bigcirc about 8.6 seconds \bigcirc about 12.6 seconds \bigcirc about 23.1 seconds \bigcirc about 3.7 seconds

Explanation:

Step1: Recall the free - fall formula

The height \(h\) of an object in free - fall (dropped, initial velocity \(v_0 = 0\)) is given by the formula \(h=- 16t^{2}+h_0\), where \(h_0\) is the initial height, \(t\) is the time in seconds, and the acceleration due to gravity near the surface of the Earth is approximated by the coefficient of \(t^{2}\) (in feet - second units, the formula is \(h=-16t^{2}+h_0\)). We know that \(h = 0\) (when the tool hits the water) and \(h_0=220\) feet. So we set up the equation \(0=-16t^{2}+220\).

Step2: Solve for \(t\)

First, we rearrange the equation:
\(16t^{2}=220\)
Then, divide both sides by 16:
\(t^{2}=\frac{220}{16}=\frac{55}{4} = 13.75\)
Next, take the square root of both sides. Since time \(t>0\), we have \(t=\sqrt{13.75}\approx3.7\) seconds.

Answer:

about 3.7 seconds