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 motion - 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.
Given $v_{0}=160$ ft/sec and $h_{0} = 45$ ft. So the equation is $h(t)=-16t^{2}+160t + 45$.

Step2: 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 object reaches its maximum height is given by the formula $t=-\frac{b}{2a}$.
Substitute $a=-16$ and $b = 160$ into the formula: $t=-\frac{160}{2\times(-16)}=\frac{160}{32}=5$ seconds.

Step3: 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) $h(t)=-16t^{2}+160t + 45$
(b) The maximum height is 445 ft and it takes 5 seconds to reach that height.