Question

a baseball throwing machine throws a baseball straight up with an initial velocity of 160 ft/sec from a height of 45 ft. (a) find an equation that models the height, h, of the ball t seconds after it is thrown. (b) what is the maximum height the baseball will reach? how many seconds will it take to reach that height? (a) the equation is

Explanation:

Step1: Recall the kinematic - height formula

The general formula for the height \(h(t)\) of an object in vertical - motion under the influence of gravity is \(h(t)=-16t^{2}+v_{0}t + h_{0}\), where \(v_{0}\) is the initial velocity and \(h_{0}\) is the initial height.

Step2: Identify the given values

We are given that \(v_{0}=160\) ft/sec and \(h_{0} = 45\) ft.

Step3: Write the height - equation

Substitute \(v_{0}=160\) and \(h_{0}=45\) into the formula \(h(t)=-16t^{2}+v_{0}t + h_{0}\), we get \(h(t)=-16t^{2}+160t + 45\).

Step4: For part (b), find the time to reach the maximum height

The function \(h(t)=-16t^{2}+160t + 45\) is a quadratic function in the form \(y = ax^{2}+bx + c\) with \(a=-16\), \(b = 160\), and \(c = 45\). The time \(t\) at which the maximum of a quadratic function \(y=ax^{2}+bx + c\) occurs is given by \(t=-\frac{b}{2a}\).
Substitute \(a=-16\) and \(b = 160\) into \(t =-\frac{b}{2a}\), we have \(t=-\frac{160}{2\times(-16)}=\frac{160}{32}=5\) seconds.

Step5: Find the maximum height

Substitute \(t = 5\) into the height - function \(h(t)=-16t^{2}+160t + 45\).
\(h(5)=-16\times5^{2}+160\times5 + 45=-16\times25+800 + 45=-400+800 + 45=445\) ft.

Answer:

(a) The equation is \(h(t)=-16t^{2}+160t + 45\)
(b) The maximum height is 445 ft and it takes 5 seconds to reach that height.