find max height, range of the projectile

Question

Starting at (0,0) a projectile travels along the path y= -(1/256)x^2 +4x where x is in meters. 

Find:

a. the maximum height

b. the range of the projectile

c. sketch the path of the projectile and determine the coordinates of the vertex and focus

Answer 1

To find the maximum height, we need to find the vertex.  There are two ways to do this.  We can write the function of the path in vertex form, or we can use the formula

 

Vertex = (h, k)

 

where:

 

h = -b / (2a)

 

k = (4ac - b²) / 4a

 

 

a = -1/256

b = 4

c = 0

 

 

Plug in these values to get your vertex.  The value of the k coordinate is your maximum height.

 

 

b)

 

The range is the set of y values where the function exists.  But in this case, x is the range because we want to know how far the projectile travels before it hits the ground.  To do this we find the x-intercepts.  The x-axis indicates ground level because the height is zero at this point.

 

 

To find the x-intercepts, we set y=0.

 

0 = -(1/256)x² + 4x

 

0 = x((-1/256)x + 4)

 

 

Set the factors equal to zero.

 

x = 0             and         (-1/256)x + 4 = 0

 

                                   (-1/256)x = -4

 

                                    x = -4(-256)

                                    x = 1024

 

 

The range of the projectile is    0 ≤ x ≤ 1024

 

 

This means the projectile travels 1024 meters before it hits the ground.

 

 

d) 

 

Now  sketch the graph using the intercepts and vertex.  Keep in mind the y-intercept is (0, 0).