Question

1. a cannon shoots a ball into the air at a football stadium. the path of the football is modeled by the equation \\( y = -20x^2 + 120x - 70 \\). (a) when will the ball reach its maximum height? (b) what is the maximum height (feet)? show work.

(a) _____________
(b) _____________

Explanation:

Part (A)

Step1: Identify the quadratic form

The equation of the ball's path is a quadratic function in the form \( y = ax^2 + bx + c \), where \( a = -20 \), \( b = 120 \), and \( c = -70 \). For a quadratic function, the time (x - value) at which the maximum height occurs is given by the formula \( x = -\frac{b}{2a} \).

Step2: Substitute the values of a and b

Substitute \( a = -20 \) and \( b = 120 \) into the formula:
\( x = -\frac{120}{2\times(-20)} \)

Step3: Simplify the expression

First, calculate the denominator: \( 2\times(-20)= -40 \)
Then, \( x = -\frac{120}{-40} = 3 \)

Part (B)

Step1: Substitute x = 3 into the original equation

Now that we know the time at which the maximum height occurs is \( x = 3 \), we substitute this value into the equation \( y = -20x^2 + 120x - 70 \) to find the maximum height (y - value).

Step2: Calculate each term

• For the first term: \( -20\times(3)^2 = -20\times9 = -180 \)
• For the second term: \( 120\times3 = 360 \)
• The third term is \( -70 \)

Step3: Combine the terms

\( y = -180 + 360 - 70 \)
\( y = ( -180 + 360 ) - 70 \)
\( y = 180 - 70 = 110 \)

Answer:

(A):
The ball will reach its maximum height at \( x = 3 \) units of time (the unit of time is not specified in the problem, but typically for such projectile motion problems, it could be seconds, etc.).