Question

c. how many feet above the ground is the ball when you hit the bossaball? the ball is 6.5 feet above the ground when you hit the bossaball. explain your reasoning.

Explanation:

When hitting the bossaball, the time \( t = 0 \) (since that's the moment of impact). If we assume the height \( h(t) \) of the ball is modeled by a function (e.g., a quadratic function for projectile motion, \( h(t)=at^{2}+bt + c \)), at \( t = 0 \), \( h(0)=c \). Here, the value \( 6.5 \) is the initial height (when \( t = 0 \)), so that's the height when the ball is hit.

Answer:

The ball is 6.5 feet above the ground when hit because at the moment of hitting (time \( t = 0 \) if we model the ball's height as a function of time), the initial height (the constant term in a height - time function, like a projectile motion model \( h(t)=at^{2}+bt + c \)) is 6.5 feet.