Question

Need help with quadratic equation and projectile formula!
In order to synchronize the neighboring fountain head to begin as soon as this one returns to the ground, you need to determine the time the water from this motor is in the air.
Using your graphing tool, find the x-intercepts for the chosen motor.
Save a copy of this motor’s final graph and paste it here or attach it to the assignment.

I'm using Geogebra to graph and the coordinates I have for the chosen motor are:
(6.5625, 689.0625)

Am I supposed to use the quadratic formula first or the projectile formula?

Answer 1

You don't need you use any formula. You have the intercepts already!

Answer 2

what is your quadratic formula ? and what is the projectile formula ?

Answer 3

The quadratic formula is x= -b ±√b^2 - 4ac/2a and the projectile formula is 0= -16^2+vx+0

I have the intercepts, but I still need to make the parabola.

Answer 4

ok, the projectile formula is
y= -16x^2 +vx

y is the height. what is x ? time ?

Answer 5

I believe so.

Answer 6

I think I'm supposed to put one of the numbers that I have for V - I know my teacher said that the V was important.

Answer 7

so the water is on the ground when y=0 or when
-16x^2 +vx =0
this happens at x=0 (at the start)
and at
-16x+ v=0 or x= v/16 (when it hits the ground coming down)

this second x-intercept is the time the water was in the air.

Answer 8

I don't know what this means
he coordinates I have for the chosen motor are:
(6.5625, 689.0625)

Answer 9

We had to graph three different equations and make a parabola. The equation for that was:
y = -16x^2 + 210x + 0
That was the equation I had to choose to use in this part because when I graphed it, it said the x intercepts were (6.5625, 689.0625)

Answer 10

the x-intercepts are where the parabola crosses the x-axis

the point (6.5625, 689.0625) is the vertex. It happens when x= 6.56... and the parabola reaches a max height of 689.06...