Question

1. the empire state building is 1250 ft tall. an object is thrown upwards from the top of the building at an initial velocity of 38 feet per second a. write the equation that represents the height (h) of the object that was throw with respect to time (t) b. when will the object reach its maximum height? c. what is the maximum height of the object?

Explanation:

Step1: Write the equation

The general form of the height - time equation for an object in vertical motion is \(h(t)=-16t^{2}+v_{0}t + h_{0}\), where \(v_{0}\) is the initial velocity and \(h_{0}\) is the initial height. Given \(v_{0} = 38\) ft/s and \(h_{0}=1250\) ft, the equation is \(h(t)=-16t^{2}+38t + 1250\).

Step2: Find the time at maximum height

For a quadratic function \(y = ax^{2}+bx + c\) (\(a=-16\), \(b = 38\)), the time \(t\) at the vertex (maximum for \(a<0\)) is given by \(t=-\frac{b}{2a}\). Substituting \(a=-16\) and \(b = 38\), we get \(t=-\frac{38}{2\times(-16)}=\frac{38}{32}=\frac{19}{16}\) s.

Step3: Calculate the maximum height

Substitute \(t = \frac{19}{16}\) into \(h(t)\):

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Answer:

a. \(h(t)=-16t^{2}+38t + 1250\)
b. \(t=\frac{19}{16}\text{ s}\)
c. \(h = 1272.5625\text{ ft}\)