## 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 maxim

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?

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1 week 2021-09-15T18:50:58+00:00 2 Answers 0

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

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.