How to solve this Quadratic Applications question?!?!

[Year10Student]

A javelin is thrown from a height of 2m above the ground.
It lands 60 meters away. The path of the javelin can be represented by the equation y=ax²+bx+c

a) What does x and y represent in this situation?

b)If it is known that the point (10,15) lies on the path given by the equation y=ax²+bx+c, use algebra and all the information given in a) and b) to show that: a = -2/75, b = 47/30 and c =2

c)Hence find the height of the javelin after travelling 20m horizontally

For a) Does y represent the height of the javelin, and x represent the horizontal distance traveled?
For b) I have no idea :c
For c) Really stuck!

 

[ksdhart]

For part a, I believe your interpretation is correct. For part b, think about what you know. You're given a point in the x-y plane and told it lies on the path given by the equation. What does each number in the point represent? What do you think it means that this point lies on the path of the equation? After you make appropriate substitutions, what does the equation look like now? Also remember that you have a second equation you can form using the information given in the introduction. Then use your algebra skills to solve for a, b, and c. They should be exactly the same as the values you were given. Once you've found the information in part b, you should easily be able to solve part c.

 

[Explorer]

a) You are correct, y represent the height (or vertical position) of the javelin, and x its horizontal position.

b) First start by substituting those values for a, b and c. You end up with:

y= -2/75 x² + 47/30 x + 2

Now to verify that the values are correct, see whether the given point (10,15) is a solution. In other words, substitute x with 10 and y with 15 and check whether the equality is satisfied.
You should end up with 15 = 15, which means that the equality is satisfied and the values given for a, b, and c are correct.

c) At this point you already know the equation of motion:

y= -2/75 x² + 47/30 x + 2

You are given the horizontal distance (20m), and you want the height. In other words, you are given a value for x, and you want to find the corresponding value for y.
All you need to do is substitute 20m for x in the equation, and solve for y.