Question

a ball is dropped from a 30-foot-tall building, meaning it has no initial velocity. write a model ( h(t) ) that represents the height of the ball from the ground, in feet, ( t ) seconds after it is dropped from the building. (2 points)
( h(t) = square t^2 + 0 t + 30 )

Explanation:

Step1: Recall free - fall formula

The general formula for the height of an object in free - fall (near the surface of the Earth) is given by \(h(t)=h_0 + v_0t-\frac{1}{2}gt^2\), where \(h_0\) is the initial height, \(v_0\) is the initial velocity, and \(g\) is the acceleration due to gravity. In the English system (feet), the acceleration due to gravity \(g = 32\) feet per second squared. But since the object is in free - fall (dropped, so \(v_0 = 0\)) and we are considering the height from the ground, the formula for the height of the ball as a function of time \(t\) (in seconds) when it is dropped from an initial height \(h_0\) is \(h(t)=h_0-\frac{1}{2}gt^2\) (because the height decreases as the ball falls).

Given that the initial height \(h_0 = 30\) feet, the initial velocity \(v_0=0\) (dropped, no initial velocity), and \(g = 32\) ft/s². So \(\frac{1}{2}g=\frac{1}{2}\times32 = 16\). Since the height is decreasing, the coefficient of \(t^2\) is negative. But in the given form \(h(t)=\square t^{2}+0t + 30\), we need to find the coefficient of \(t^2\).

The formula for the height of the ball above the ground is \(h(t)=- 16t^{2}+0t + 30\) because the ball is falling downwards, so the quadratic term has a negative coefficient (the height decreases as \(t\) increases).

Step2: Identify the coefficient

In the equation \(h(t)=at^{2}+bt + c\), where \(a\) is the coefficient of \(t^2\), \(b\) is the coefficient of \(t\), and \(c\) is the constant term. Here, \(b = 0\) (no initial velocity), \(c = 30\) (initial height), and \(a=- 16\) (because of the acceleration due to gravity causing the height to decrease quadratically with time).

Answer:

\(h(t)=\boxed{-16}t^{2}+0t + 30\)