Question

A dolphin jumps out of the water. The function h=-16t^2+26t models the height (in feet) of the dolphin after t seconds. After how many seconds is the dolphin at a height of 5 feet.

Answer 1

The dolphin is five feet in the air after about 0.22 and 1.40 seconds.

Explanation:

The height h (in feet) of the dolphin as it jumped out of the water after t seconds is given by the function:

`h(t)=-16t^2+26t`

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:

`5=-16t^2+26t`

Solve for t. Rearrange:

`16t^2-26t+5=0`

We can use the quadratic formula, given by:

`\displaystyle t=(-b\pm√(b^2-4ac))/(2a)`

In this case, a = 16, b = -26, and c = 5.

Substitute:

`\displaystyle t=(-(-26)\pm√((-26)^2-4(16)(5)))/(2(16))`

Evaluate:

`\displaystyle t=(26\pm√(356))/(32)`

Simplify the square root:

`√(356)=√(2\cdot 2\cdot 89)=2√(89)`

Hence:

`\displaystyle t=(26\pm2√(89))/(32)`

Simplify:

`\displaystyle t=(13\pm√(89))/(16)`

Hence, our solutions are:

`\displaystyle t=(13-√(89))/(16)\approx 0.22\text{ and } t=(13+√(89))/(16)\approx1.40`

The dolphin is five feet in the air after about 0.22 and 1.40 seconds.