If a ball is thrown directly upward with a velocity of 50ft/s, it's height (in feet) after t seconds is given by y=50t-16t^2 what is the maximum height attained by the ball

39.0625 ft

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 50

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

We have x = - 50/[2 (-16) ]

x = - 50/-32

x = 25/16

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

y = 50 (25/16) - 16 (25/16) ²

y = 1250/16 - 16 (625/256)

y = 1250/16 - 625/16

y = 625/16

y = 39.0625