Aball is thrown vertically in the air with an initial velocity of 60 feet per second from an initial height of 6 feet. the equation describing the height of the ball at any time is s = -16t^2 +60t+6.

1. how long did it take for the ball to reach its height point?
2. how high did the ball go?

Step-by-step explanation:

The problem can be solved by putting the equation into vertex form.

  s - 6 = -16(t^2 -15/4t)

  s -6 -16(15/8)^2 = -16(t^2 -15/4t +(15/8)^2) . . . . add the square of half the x-coefficient inside parentheses and equivalent amount on the other side of the equation

  s -62.25 = -16(t -15/8)^2 . . . . write as a square, simplify the constant

  s = -16(t -1.875)^2 +62.25 . . . . put in vertex form

The vertex of the downward-opening parabola is at (t, s) = (1.875, 62.25).

1. It took 1.875 seconds for the ball to reach its maximum height.

2. The ball went up to 62.25 feet.

y - y1 = m(x- x1)

y - (-5) = 2/3(x -0)

y + 5 = 2/3x

y = 2/3x - 5

n

step-by-step explanation: