Question

The height of a ball above the ground as a function of time is given by the functionℎ()=−32^2+8+3where h is the height of the ball in feet and t is the time in seconds. When does the ball hit the ground? Round to the nearest thousandth (3 places past the decimal).

Answer 1

The height of the ball is modeled by

$h(t)=-32t^2+8t+3$

When the ball hit the ground, the value of h(t) becomes 0.

Now, let us solve this equation.

$\begin{gathered} h(t)=0 \\ -32t^2+8t+3=0 \\ 32t^2-8t-3=0 \\ t=\frac{8\pm\sqrt[]{64+384}}{64} \\ =0.456,-0.206 \end{gathered}$

So, the ball hit the ground after 0.456 seconds