Aphysics student stands at the top of a hill that has an elevation of 37 meters he throws a rock and it goes up into the air and then falls back past him and lands on ground below .the path of the rock can be modeled by the equation y =-0.02x^2+0.8x+37 where x is the horizontal distance, in meters ,from the starting point on the top of the hill and y is the height, in meters, of the rock above the ground. how far horizontally from its starting point will the rock land? round your answer to the nearest hundredth.

The equation you are given models the path of the rock. x is the horizontal distance from the start. y is the height of the rock from the ground.

You are asked to find the horizontal distance (x) when the rock is on the ground. When the rock is on the ground its height (y) will be 0. It is 0 meters off the ground.

That means the question is asking us to solve for x when y=0. Let's substitute 0 for y in the given equation.



This is a quadratic equation and can be solved using the quadratic formula. The quadratic formula is:


The "plusminus" I wrote is usually written as the plus symbol over the minus symbol. More on that a little later.

We need to identify a, b and c.
a is the coefficient of the squared term (of ). That is, .
b is the coefficient of the linear term (x). That is, b=.8
c is the constant (the number that is by itself). That is, c=37

We use these values in the quadratic formula to find x as follows:





Now we deal with the "plusminus" by splitting the equation in two. In one you use "plus' and in the other "minus".


and


Since we are looking for a distance the answer cannot be negative so we disregard the negative answer. The horizontal distance (rounded to 2 decimal places as requested) is 15.75 meters.