A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the function h (t) = 52t - 16t^2. What is the maximum height that the ball will reach?

Do not round your answe

42.25 feet.

Step-by-step explanation:

The maximum height can be found by converting to vertex form:

h (t) = 52t - 16t^2

h (t) = - 16 (t^2 - 3.25t)

h (t) = - 16 [ (t - 1.625) ^2 - 2.640625 ]

= - 16 (t - 1.625 ^2) + 42.25

Maximum height = 42.25 feet.

Another method of solving this is by using Calculus:

h (t) = 52t - 16t^2

Finding the derivative:

h' (t) = 52 - 32t

This = zero for a maximum/minimum value.

52 - 32t = 0

t = 1.625 seconds at maximum height.

It is a maximum because the path is a parabola which opens downwards. we know this because of the negative coefficient of x^2.

Substituting in the original formula:h (t) = 52 (1.625) - 16 (1.625) ^2

= 42.25 feet.

42.25 feet

Step-by-step explanation:

The height function is a parabola. The maximum value of a negative parabola is at the vertex, which can be found with:

x = - b/2a

where a and b are the coefficients in y = ax² + bx + c.

Here, we have y = - 16t² + 52t. So a = - 16 and b = 52. The vertex is at:

t = - 52 / (2*-16)

t = 13/8

Evaluating the function:

h (13/8) = - 16 (13/8) ² + 52 (13/8)

h (13/8) = - 169/4 + 169/2

h (13/8) = 169/4

h (13/8) = 42.25