Question

during a baseball game, a player's ball hit into the outfield is modeled by the equation h=-16t^2 + 65t. A bird flying across the field follows the path modeled by the equation h=8t+20. let t be the time since the ball is hit and h be the height. what do the intersection points of the equations represent?

Answer 1

Answer:

Intersection points are

(0.39,0) and (3.16,0)

Step-by-step explanation:

The equation of the outfield is $h=-16t^2 + 65t$

The path equation is $h=8t+20$

To find intersection of two equations , we make them equal and solve for t

$-16t^2+65t= 8t+20$

Subtract 8t from both sides

$-16t^2+57t=20$

Subtract 20 on both sides

$-16t^2+57t-20=0$

Divide whole equation by -1

$16t^2-57t+20=0$

Apply quadratic formula to solve for t

$t=\frac{-b+-\sqrt{b^2-4ac}}{2a}$

a= 16, b=-57 and c=20

$t=\frac{57+-\sqrt{(-57)^2-4(16)(20)}}{2(16)}$

$t=\frac{57+-\sqrt{1969}}{32}$

$t=\frac{57+\sqrt{1969}}{32}=3.16$

$t=\frac{57-\sqrt{1969}}{32}=0.39$

Answer 2

The intersections of the equations represent the times when the bird and the ball are at the same height.

8t+20=-16t^2+65t

16t^2-57t+20=0, using the quadratic equation for expediency:

t=(57±√4529)/32 and since we know t>0

t≈3.88 seconds  (at a approximate height of 51 feet)