A cannonball is launched from a cannon on a wall. If the path is traced on piece of paper so that the cannon is situated on the y axis, the equation of the path is y= -1/600x^2 + 1/2x + 20. How high above the ground is the cannon.
20 (the constant term), which is the value when x = 0.
To determine how high above the ground the cannon is, we need to find the y-coordinate when the cannonball is at the ground. In other words, we want to find the y-intercept of the path.
The equation of the path is given as y = -1/600x^2 + 1/2x + 20.
To find the y-intercept, we set x = 0 in the equation and solve for y:
y = -1/600(0)^2 + 1/2(0) + 20
y = 0 + 0 + 20
y = 20
Therefore, the y-coordinate of the y-intercept is 20.
This means that the cannon is situated 20 units above the ground.