Question

A ball is launched from a platform. The equation for the ball's height h at time t seconds after launch is h(t)= -16t^2+153.6t +393.12, where h is in feet. What is the initial height of the ball before launch?

Answer 1

Answer:

The initial height of the ball before launch is 393.12 feet.

Step-by-step explanation:

The ball's height ($h$), in feet, as a function of time ($t$), in seconds, is represented by $h(t) = -16\cdot t^{2}+153.6\cdot t + 393.12$. The initial height of the ball before launch is the height of the ball evaluated at $t = 0\,s$:

$h(0) = -16\cdot (0)^{2}+153.6\cdot (0) + 393.12$

$h(0) = 393.12\,ft$

The initial height of the ball before launch is 393.12 feet.