Question

I need some help on the Math equation in the picture I took.

Answer 1

4.

$h(t)=-t^2+32t+48$

a)

We need to find the roots of the equation, so:

$\begin{gathered} h(t)=0 \\ so\colon \\ -t^2+32t+48=0 \end{gathered}$

Solve for t using quadratic formula:

$\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ _{\text{ }}where\colon \\ a=-1 \\ b=32 \\ c=48 \\ so\colon \\ t=\frac{-32\pm\sqrt[]{1024+192}}{-2} \\ so\colon \\ t1=\frac{-32+8\sqrt[]{19}}{-2} \\ t2=\frac{-32-8\sqrt[]{19}}{-2} \end{gathered}$

So, the total time T is:

$T=t2=33.436$

Answer:

33.436 s

b)

We need to evaluate the function for t = 1:

$\begin{gathered} h(1)=-(1)^2+32(1)+48 \\ h(1)=79ft \end{gathered}$

Answer:

79ft

c)

The initial height is the height of the ball for t = 0, so:

$\begin{gathered} t=0 \\ h(0)=-0^2+32(0)+48 \\ h(0)=48ft \end{gathered}$

Answer:

48ft