Question

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

Answer 1

42.25 feet.

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.

Answer 2

42.25 feet

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