Question

a softball pitcher tosses a ball straight upward from a height of 3 feet above the ground with an initial velocity of 50 feet per second. the acceleration due to gravity is $-16$ ft/s². which quadratic equation represents the height of the ball above the ground after t seconds?
$h(t) = at^2 + vt + h_0$

• $h(t) = 50t^2 - 16t + 3$
• $h(t) = -16t^2 + 50t + 3$
• $3 = -16t^2 + 50t + h_0$
• $3 = 50t^2 - 16t + h_0$

Explanation:

Step1: Identify the values of a, v, and \( h_0 \)

In the formula \( h(t) = at^2 + vt + h_0 \), \( a \) is the acceleration, \( v \) is the initial velocity, and \( h_0 \) is the initial height.
Given: \( a=-16 \, \text{ft/s}^2 \) (acceleration due to gravity), \( v = 50 \, \text{ft/s} \) (initial velocity), \( h_0=3 \, \text{ft} \) (initial height).

Step2: Substitute the values into the 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:

\( h(t) = -16t^2 + 50t + 3 \) (the second option: \( h(t) = -16t^2 + 50t + 3 \))