# y= -16x^2+48x+6 ( y is the height and x is time) if the ball is initally thrown from 6 ft how many seconds after the ball is thrown will it again be 6ft above ground?what is the max height in ft...

y= -16x^2+48x+6 ( y is the height and x is time) if the ball is initally thrown from 6 ft how many seconds after the ball is thrown will it again be 6ft above ground?

what is the max height in ft the ball reaches? ... how would i sove this proublem any help at all is greatly appreciated!!(:

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**First question:**

The easiest way to do this is to recognize that "y," your height, is 6 ft in the solution, so you get the following equation where you'll solve for "x."

6 = -16x^2 + 48x + 6

Now subtract 6 from both sides

0 = -16x^2 + 48x

Now you can factor out the x on the right side:

0 = x(-16x + 48)

Now, we have two ways this equation can be zero because when you multiply two numbers and the result is zero, at least one of the numbers is zero:

1) x = 0

OR

2) -16x + 48 = 0

The first case is the trivial case: the height is 6 at x = 0 because that's where the ball started at time 0! We don't care about this case, so we'll move on and solve for "x" in the second case:

-16x + 48 = 0

Subtract 48:

-16x = -48

Divide by -16 to get x:

**x = 3**

3 seconds sounds like a reasonable answer, so we just put 3 back into our original equation to make sure it's correct:

6 = -16(3^2) + 48(3) + 6

6 = -144 + 144 + 6

6 = 6

Looks like we're good! We can now say for sure that our answer is **x = 3 seconds**.

**Second Question:**

We know the trajectory of the ball is a parabola because the equation is a polynomial with x^2 being the highest power of x. Because it's a parabola, we can say that the maximum height will be at the time that is the **average of any two times where the ball is at the same height**. Put another way, if I know our ball is at 6 ft at times x = 0 and x = 3, I can say the time that the ball is at the maximum height is right at the midpoint of those two times:

x(maximum height) = (0 + 3)/2 = **1.5 seconds**

Now to find the maximum height the ball reaches, we simply put in the time we found (x = 1.5 seconds):

y = -16(1.5^2) + 48(1.5) + 6

Now just simplify and figure out your height:

y = -16(2.25) + 48(1.5) + 6

y = -36 + 72 + 6

**y = 42 ft**

**Therefore, the maximum height the ball reaches is 42 ft at 1.5 seconds into its trip.**

## y= -16x^2+48x+6 ( y is the height and x is time) if the ball is initally thrown from 6 ft how many seconds after the ball is thrown will it again be 6ft above ground?

We have to find when y = 6.

`6 = -16x^2+48x+6`

`0 = -16x^2 + 48x`

`0=-16x(x-3)`

`0=x(x-3)`

So when x = 0 or x = 3 the ball is six feet off the ground.