Question

Sam launched a ball vertically upward from the ground at the same time a bird was on a linear path toward the ground. The equation that models the approximate position, in meters, of the ball is p = -5t^2 + 18t, and the equation that models the path of the bird is p = -2t + 15. At what position should the ball first cross the path of the bird?

a) 3 meters
b) 0 meters
c) 1 meter
d) 13 meters

Answer 1

The ball first crosses the path of the bird at 13 meters above the ground, which is determined by setting the equations for the positions of the ball and bird equal to each other and solving for time.

Step-by-step explanation:

To determine at what position the ball first crosses the path of the bird, we need to find the point at which their positions as functions of time are equal. Given that the position of the ball is modeled by the equation p = -5t^2 + 18t and the bird's path is modeled by the equation p = -2t + 15, we can set the two equations equal to each other to solve for the time t when their positions are the same:

-5t^2 + 18t = -2t + 15

This equation simplifies to:

-5t^2 + 20t - 15 = 0

Factoring the quadratic equation we get:

-5(t^2 - 4t + 3) = 0

-5(t - 3)(t - 1) = 0

The solutions for t are t = 3 seconds and t = 1 second. Since we are looking for the first time they cross, we take the smaller positive value t = 1 second. We then substitute t = 1 second into either of the given equations to find the position. Substituting into the bird's equation:

p = -2(1) + 15 = 13 meters

So, the ball first crosses the path of the bird at 13 meters.