Respuesta :
Given the function:
[tex]h(t)=-16t^2+8t+48[/tex]Where h(t) is the height of the diver above the surface of the water and t is the time.
Let's find how long it takes the diver to hit the water.
When the diver hits the water, the height h(t) = 0.
Now substitute 0 for h(t) and solve for the time t.
We have:
[tex]0=-16t^2+8t+48[/tex]Rearrange the equation:
[tex]-16t^2+8t+48=0[/tex]Solve for t.
Let's factor the expression by the left.
Factor 8 out of all terms:
[tex]8(-2t^2+t+6)=0[/tex]Now, factor by grouping.
Rewrite the middle term as a sum of two terms whose product is the product of the first term and the last term:
[tex]\begin{gathered} 8(-2t^2+4t-3t+6)=0 \\ \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} 8((-2t^2+4t)(-3t+6))=0 \\ \\ 8(2t(-t+2)+3(-t+2))=0 \\ \\ 8(2t+3)(-t+2)=0 \end{gathered}[/tex]Hence, we have the factors:
[tex]\begin{gathered} 2t+3=0 \\ -t+2=0 \end{gathered}[/tex]Solve each factor for t:
[tex]\begin{gathered} 2t+3=0 \\ \text{ Subtract 3 from both sides:} \\ 2t=-3 \\ \text{ Divide both sides by 2:} \\ \frac{2t}{2}=-\frac{3}{2} \\ t=-\frac{3}{2} \\ \\ \\ -t+2=0 \\ t=2 \end{gathered}[/tex]Hence, we have the solutions:
t = -3/2
t = 2
The time cannot be negative, so let's take the positive value.
Therefore, the will take 2 seconds for the diver to hit the water.
ANSWER:
2 seconds.
