## anonymous 5 years ago a baseball is thrown from height of 2meters and caught at the same height 38meters away. During its parabolic path, it reaches a maximum height of 16meters. Find the equation in std form which relates the height of the ball to the horizontal distance that it has traveled.

From the given information we know that $$(0,2)$$ is the $$y$$-intercpet. And also $$(38,2)$$ is another point on the parabola. The line of symmetry is in the midpint between $$x=0$$ and $$x=38$$. So the line $$x=19$$ is the line of symmetry. Since the maximum height is 16 meters, then $$(19,16)$$ is the vertex. Hence in the vertex form the parabola is of the form $$y=a(x-19)^2+16$$ Plugin $$(0,2)$$ we have $$2=19^2a+16$$ $$a=-\frac{14}{361}$$. Hence $$y=-\frac{14}{361}(x-19)^2+16$$ $$y=-\frac{14}{361}x^2+\frac{28}{19}x+2$$