Question

2.6.26 Deon throws a ball into the air. The height h of the ball, in meters, at time t seconds is modeled by the function h(t) = -5t^2+t+4. When will the ball hit the ground?

Answer 1

1 second after Deon throws it

1) Since the ball goes in the air as described by h(t) = -5t²+t +4, then we can visualize roughly this way:

2) Then we can assume that the moment the ball hits the ground, is one of the roots of that equation. So let's solve it:

$\begin{gathered} h(t)=-5t^2+t+4 \\ -5t^2+t+4=0 \\ \Delta=b^2-4ac\text{ }\Rightarrow\Delta=(1)^2-4(-5)(4)\Rightarrow\Delta=81 \\ x=\frac{-b\pm\sqrt[]{\Delta}}{2a}\Rightarrow x=\frac{-1\pm9}{2(-5)}=1\text{ and -4/5} \end{gathered}$

3) So since the negative root of it does not matter to us, since there's no negative time. The ball will hit the ground 1 second after Deon throws it.