Question

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.

Answer 1

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
\(y=-\frac{14}{361}(x-19)^2+16\)
\(y=-\frac{14}{361}x^2+\frac{28}{19}x+2\)