# 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!!(:

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

...

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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.

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## 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.

Approved by eNotes Editorial Team