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

Question

A ball is thrown from a height of 6 Feet.

1.)How many seconds after the ball is thrown will it again be at 6 feet above ground?

2.)What is the maxmum height, in feet, that the ball reaches?

Answer 1

1). Solve: 6 = -16x^2 + 48x + 6

0 = -16x^2 + 48x (subtract 6 from each side)
0 = -16x(x -3) (factor)
x = 0, 3 (zero-product property)
The answer is 3 seconds.

2) The vertex occurs at x = -b/(2a) = -48/(2*-16) = 1.5

Max height = -16*(1.5)^2 + 48(1.5) + 6 = 42 ft.

Answer 2

6 = -16x^2 + 48x + 6
=> x^2 = 3x
=> x = 3.
If x is the time then the answer is 3.
max. height will be achieved in between., i.e at x = 1.5
=> y = -16 * 3/2 * 3/2 + 48 * 3/2 + 6 = -36 + 72 + 6 = 42

Answer 3

Your problem lacks data such as the initial velocity and angle with the horizontal.