A ball is thrown upward from the roof of a 25m building. The ball reaches a height of 45m above the ground after 2s and hits the ground 5s after being thrown. What is the equation for this function?

1 Answer
Apr 29, 2018

# y = -5(x-2)^2+45 {0 ≤ x ≤ 5}#


Given that the ball reaches a maximum height of 45 metres after 2 seconds, your vertex is going to be at (2,45), given that the distance is dependent on time. You know that the graph has an x-intercept at (5,0) because that’s when the ball hits the ground. So therefore the equation in vertex form would be:

# y = a(x - 2)^2 + 45#

But you’re not done yet, because you need to find the vertical stretch factor, #a#. Use your x-intercept to do this part:





Your new function is:

#y = -5(x-2)^2+45#

But you’re STILL not done. The current graph doesn’t completely make sense, because you can go backwards in time (unless you somehow got a hold of the Eye of Agamodo, or the Delorean Time Machine), and there’s absolutely no way that your ball can punch through the ground to go negative meters (unless it’s made from vibranium, maybe).

To resolve these problems, you have to restrict the domain of the function (the range of acceptable x-values).

Your final function is:

#y = -5(x-2)^2 +45 ; {x | 0 ≤ x ≤ 5; x in R}#

Depending on what level of math you’re in, you may or may not be required to include the domain restriction.

Hope that helps!