Ball kicked into the air with an initial upward velocity of 60 ft/s. it’s height h in feet after t seconds is given by the function h=-16t^2 +60t+5. what is the balls maximum height? how long did it take for the ball to hit the ground? how long did it take for the ball to reach its maximum height? how high is the ball one second after it is kicked?

Max height: 61.25 feet

Max height reached at 15/8 seconds, or 1.875 seconds

How long to hit the ground: (15 + 7√5)/8  or about 3.83 seconds

How high after 1 second:  49 feet

Step-by-step explanation:

Since the function is a quadratic representing height, and the coefficient of the t² is negative, the vertex of the parabola will be the maximum height achieved by the ball.  

The general form for a quadratic equation is ax² + bx + c,

 here a is -16, and b is 60

To find the x coordinate of the vertex, use   x = -b/(2a)

  We have x = -60/[2(-16)]

                x = -60/-32

                         x = 15/8

So at 15/8 seconds, the ball reaches is maximum height

Now plug that into the equation to find the y value, which will be the height...

y = -16(15/8)² + 60(15/8) + 5  

      y = -16(225/64) + 900/8 + 5

     y = -225/4 +  450/4 + 20/4

       y =  245/4

           y = 61.25 feet

To find out how long the ball was in flight, solve the equation...

0 = -16t² + 60t + 5

    Use quadratic equation...

    x = -60/-32 ± √[60² - 4(-16)(5)]/-32

     x = 15/8 ± (√3920)/-32

      x = 15/8 ± (28√5)/-32

          x = 15/8 ± (-7√5)/8

  so x = (15 - 7√5)/8   and  (15 + 7√5)/8

     (15 - 7√5)/8 is negative, and we're talking about time, so this answer is ignored.  

(15 + 7√5)/8 seconds is when the ball hits the ground

To find out out how high the ball was after 1 second, plug 1 in for x and simplify

h(1) = -16(1²) + 60(1) + 5

 h(1) = -16 + 60 + 5

  h(1) = 49

p> q if you calculate p is100 q is 79

this answer is !