Question

a softball pitcher throws a softball to a catcher behind home plate. the softball is 3 feet above the ground when it leaves the pitchers hand at a velocity of 50 feet per second. if the softballs acceleration is -16 ft/s², which quadratic equation models the situation correctly?
h(t)=at² + vt+h₀
○ h(t)=50t² - 16t + 3
○ h(t)= - 16t²+50t + 3
○ 3=-16t² + 50t+h₀
○ 3 = 50t²-16t+h₀

Explanation:

Step1: Identify the values

The general height - time formula is $h(t)=at^{2}+vt + h_{0}$, where $a$ is the acceleration, $v$ is the initial velocity and $h_{0}$ is the initial height. Given $a=- 16$ (acceleration), $v = 50$ (initial velocity) and $h_{0}=3$ (initial height).

Step2: Substitute values into formula

Substitute $a=-16$, $v = 50$ and $h_{0}=3$ into $h(t)=at^{2}+vt + h_{0}$, we get $h(t)=-16t^{2}+50t + 3$.

Answer:

$h(t)=-16t^{2}+50t + 3$ (corresponding to the second option in the multiple - choice list)