Question

a ball is thrown straight upward. a position function ( s(t)=\frac{1}{2} g t^{2}+v_{0} t+s_{0} ) can be used to describe the balls height, ( s(t) ), in foot, after ( t ) seconds. a graph of the balls height is shown to the right. complete parts a through c below. a. use the points labeled in the graph to find the values of ( g, v_{0} ), and ( s_{0} ). solve the system of linear equations involving ( g, v_{0} ), and ( s_{0} ) using matrices. ( g=-32 ) ( v_{0}=52 ) ( s_{0}=0 ) b. find and interpret ( s(3.25) ) ( s(3.25)= ) (simplify your answer.)

Explanation:

Step1: Substitute values into the position function

We know \(g = - 32\), \(v_0=52\), \(s_0 = 0\), and the position function is \(s(t)=\frac{1}{2}gt^{2}+v_0t + s_0\). Substitute \(t = 3.25\) into the function:

$$s(3.25)=\frac{1}{2}\times(-32)\times(3.25)^{2}+52\times3.25+0$$

Step2: Calculate each term

First, calculate \(\frac{1}{2}\times(-32)\times(3.25)^{2}\):

$$
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$$

Second, calculate \(52\times3.25 = 169\)

Step3: Find the value of \(s(3.25)\)

$$s(3.25)=-169 + 169=0$$

Answer:

\(s(3.25)=0\). This means that at \(t = 3.25\) seconds, the height of the ball is \(0\) feet, which implies the ball has returned to the ground.