Answer:
The dolphin is five feet in the air after about 0.22 and 1.40 seconds.
Step-by-step explanation:
The height h (in feet) of the dolphin as it jumped out of the water after t seconds is given by the function:
[tex]h(t)=-16t^2+26t[/tex]
We want to determine the time(s) when the dolphin is five feet in the air.
Since the dolphin is five feet in the air, h(t) = 5:
[tex]5=-16t^2+26t[/tex]
Solve for t. Rearrange:
[tex]16t^2-26t+5=0[/tex]
We can use the quadratic formula, given by:
[tex]\displaystyle t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In this case, a = 16, b = -26, and c = 5.
Substitute:
[tex]\displaystyle t=\frac{-(-26)\pm\sqrt{(-26)^2-4(16)(5)}}{2(16)}[/tex]
Evaluate:
[tex]\displaystyle t=\frac{26\pm\sqrt{356}}{32}[/tex]
Simplify the square root:
[tex]\sqrt{356}=\sqrt{2\cdot 2\cdot 89}=2\sqrt{89}[/tex]
Hence:
[tex]\displaystyle t=\frac{26\pm2\sqrt{89}}{32}[/tex]
Simplify:
[tex]\displaystyle t=\frac{13\pm\sqrt{89}}{16}[/tex]
Hence, our solutions are:
[tex]\displaystyle t=\frac{13-\sqrt{89}}{16}\approx 0.22\text{ and } t=\frac{13+\sqrt{89}}{16}\approx1.40[/tex]
The dolphin is five feet in the air after about 0.22 and 1.40 seconds.
