Question

Quarterback Joe Tarkenton can throw a football downfield at a speed of 60 feet per second. When releasing the ball, his arm is 6 feet above ground. The height of the ball after seconds is given by the function: h(t)=−16^2+60+6.

What is the maximum height of the football?

If the ball is not caught, how many seconds will it take to reach the ground after being thrown?


A: maximum height=3.58 feet; time=62.25 seconds

B: maximum height=62.52 feet; time=3.58 seconds

C: maximum height=62.25 feet; time=3.85 seconds

D: maximum height=3.85 feet; time=62.52 seconds

Answer 1

C: maximum height=62.25 feet; time=3.85 seconds

Explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

`f(x) = ax^(2) + bx + c`

It's vertex is the point
`(x_(v), y_(v))`

In which

`x_(v) = -(b)/(2a)`

`y_(v) = -(\Delta)/(4a)`

Where

`\Delta = b^2-4ac`

If a<0, the vertex is a maximum point, that is, the maximum value happens at
`x_(v)`, and it's value is
`y_(v)`.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`.

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = a(x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\Delta))/(2*a)`

`x_(2) = (-b - √(\Delta))/(2*a)`

`\Delta = b^(2) - 4ac`

In this question:

The height of the ball, after t seconds, is given by the following equation:

`h(t) = -16t^2 + 60t + 6`

Which is a quadratic equation with
`a = -16, b = 60, c = 6`

Maximum height:

Since a < 0, we can find the maximum value of the function. We have that:

`\Delta = 60^(2) - 4(-16)(6) = 3984`

`y_(v) = -(3984)/(4(-16)) = 62.25`

The maximum height is of 62.25 feet.

Seconds to reach the ground:

`x_(1) = (-60 + √(3984))/(2*(-16)) = -0.1`

`x_(2) = (-60 - √(3984))/(2*(-16)) = 3.85`

Since time is a positive measure, 3.85 seconds.

The correct answer is given by option C.